Overview
This unit on Calculus develops the language and tools for studying change and accumulation. It begins with limits and continuity to define behaviour of functions near points, then builds the derivative as the rate of change and slope of tangent. Differentiation rules and techniques (product, quotient, chain, implicit, parametric) let students differentiate complex expressions. Applications include finding tangents and normals, increasing/decreasing tests, maxima and minima, concavity and points of inflection, and using Rolle's and the Mean Value Theorem. Taylor and Maclaurin series give polynomial approximations for smooth functions. L'Hospital's rule helps evaluate indeterminate forms. The integral is treated both as an antiderivative (indefinite integral) and as signed area (definite integral), with properties, techniques (substitution, parts, partial fractions) and the Fundamental Theorem connecting differentiation and integration. Applications of integration include area between curves and volumes of revolution. Improper integrals and basic reduction formulae extend integration to limits and repeated patterns. The unit closes with first-order differential equations (separable and linear) used to model growth and decay. Mastery of these topics is essential for physics, engineering and higher mathematics, and provides problem-solving skills valued in examinations and real-world applications.
Learning Objectives
- Explain the concepts of limit and continuity for real functions and compute limits using algebraic methods.
- Differentiate standard and composite functions using rules such as product, quotient and chain rule.
- Apply differentiation to find tangents, normals, increasing/decreasing intervals, maxima, minima and points of inflection.
- Use Rolle's theorem and the Mean Value Theorem to analyse function behaviour and justify conclusions.
- Evaluate limits of indeterminate forms using L'Hospital's rule and series expansions where appropriate.
- Integrate functions using substitution, integration by parts and partial fractions, and compute definite integrals.
- Apply the Fundamental Theorem of Calculus to connect differentiation and integration and to evaluate definite integrals.
- Use integration to compute areas between curves and volumes of solids of revolution.
- Solve first-order differential equations that are separable or linear and apply them to simple growth and decay models.
Topics in this chapter
18 topics · tap a topic title to jump straight to it.
Limits and their evaluation
Definition and intuitive picture. A limit describes the value a function approaches as the input approaches a specified point. Writing lim_{x→a} f(x) = L means that as x gets arbitrarily close to a (but not necessarily equal to a), the values f(x) get arbitrarily close to L. Limits let us study behaviour at points where a function might be undefined or where direct substitution is difficult.
Evaluation techniques — basic cases. The simplest case is direct substitution: if f is a polynomial or rational function without a zero denominator at a, substitute x=a. When substitution gives 0/0 or ∞/∞ we must do more work. Algebraic simplification includes factoring and cancelling common factors, rationalising numerator or denominator (multiply by conjugate for roots), combining fractions to one denominator and simplifying, or dividing numerator and denominator by highest power of x to handle limits at infinity.
Limits at infinity and horizontal asymptotes. To find lim_{x→∞} f(x) for rational functions, divide numerator and denominator by the highest power of x present to find the leading behaviour. If degrees are equal, limit is ratio of leading coefficients; if numerator degree smaller, limit is 0; if numerator degree larger, limit is ±∞ or does not exist (oblique/higher asymptote). Such limits show how function behaves far from origin and whether a horizontal asymptote exists.
One-sided limits and existence. The two-sided limit exists only when both left-hand limit lim_{x→a−} f(x) and right-hand limit lim_{x→a+} f(x) exist and are equal. For piecewise-defined functions or functions with jumps, one-sided limits are different; then two-sided limit does not exist. For vertical asymptotes, limits may be ±∞ from one or both sides.
Standard trigonometric and exponential limits. Certain limits appear frequently: lim_{x→0} (sin x)/x = 1 is used often for small-angle approximations and to find derivatives of trig functions. Limits involving expressions like (1 + x)^{1/x} tend to e as x→0. Knowing these standard limits simplifies many problems. When dealing with compositions, substitution (u=g(x)) can reduce to a standard limit when u→0.
Indeterminate forms and preparation for derivatives. A form like 0/0 or ∞/∞ is indeterminate because it gives no immediate information; algebraic manipulation or L'Hospital's rule (later topic) can resolve it. Limits are the foundation for the derivative: the derivative at a point is the limit of a difference quotient. Careful limit evaluation underpins the rigor of calculus.
- Find lim_{x→2} (x^2−4)/(x−2). Factor numerator to (x−2)(x+2), cancel and substitute to get 4.
- Evaluate lim_{x→0} (sin 3x)/(x) by setting u=3x, giving limit 3.
- Find lim_{x→∞} (3x^2 + x)/(x^2 − 5) by dividing by x^2 to obtain 3.
- Evaluate lim_{x→0} (1+x)^{1/x} which equals e.
- \[lim_{x→a} f(x) = L means values of f(x) approach L as x→a.\]
- \[lim_{x→0} (sin x)/x = 1\]
- \[lim_{x→0} (1+x)^{1/x} = e\]
Continuity of functions
Formal idea of continuity. A function f is continuous at x=a if the following three conditions hold: f(a) is defined, the limit lim_{x→a} f(x) exists, and the limit equals the function value: lim_{x→a} f(x) = f(a). Intuitively, the graph has no breaks, holes or jumps at that point. Continuity on an interval means being continuous at every point there.
Common continuous functions. Polynomials are continuous everywhere. Rational functions are continuous wherever the denominator is non-zero. Exponential, logarithmic and the basic trigonometric functions are continuous on their domains. Sum, difference, product and quotient (where denominator non-zero) of continuous functions are continuous. Composition of continuous functions is continuous; thus many complicated expressions remain continuous when their pieces are continuous.
Types of discontinuity explained. Removable discontinuity occurs when the limit exists but f(a) is either undefined or defined to a different value; graphically, a hole appears. By redefining f(a) we can make the function continuous. Jump discontinuity occurs when left- and right-hand limits exist but are unequal; common in step functions and signals. Infinite (essential) discontinuity arises when function value grows unbounded near a point, creating a vertical asymptote.
Testing continuity for piecewise functions. For piecewise definitions, check endpoints of pieces. Compute lim_{x→c−} f(x) and lim_{x→c+} f(x) and compare to f(c). If both one-sided limits equal f(c), the function is continuous at c. If not, describe the type of discontinuity and whether a removable fix exists.
Important theorems that require continuity. The Intermediate Value Theorem (IVT) states that a continuous function on [a,b] takes every value between f(a) and f(b); this guarantees existence of roots and is widely used to show solutions to equations. Rolle's theorem and the Mean Value Theorem require continuity on closed intervals and differentiability inside; they are essential for later topics on derivatives and extrema. Continuity also ensures integrability for many functions.
Practical checks and examples. To show continuity at a point do the limit calculation and compare to function value. For rational functions with cancellation, check for removable holes. Continuity is a basic property to check before applying tools like MVT or FTC; understanding the types helps in sketching graphs and solving applied problems.
- Show f(x)= (x^2−1)/(x−1) is removable at x=1; simplify to x+1 and define f(1)=2 for continuity.
- Identify discontinuity of f(x)=1/(x−3) at x=3 as infinite with vertical asymptote.
- Check continuity of a piecewise function where f(x)=x^2 for x≤2 and f(x)=5−x for x>2 by computing one-sided limits at x=2.
- \[f continuous at a ⇔ lim_{x→a} f(x) = f(a)\]
- Sum, product and quotient rules: if f,g continuous at a then f±g, f·g are continuous; f/g is continuous if g(a)≠0
The derivative: definition and interpretation
Definition by first principle. The derivative of f at a point a, written f'(a), is defined as the limit f'(a) = lim_{h→0} [f(a+h) − f(a)]/h provided this limit exists. This precise limit measures the instantaneous rate at which f changes with x at a. If the limit exists, f is said to be differentiable at a.
Geometric meaning. Geometrically, f'(a) is the slope of the tangent line to the graph y=f(x) at x=a. The difference quotient [f(a+h) − f(a)]/h gives the slope of the secant line through (a, f(a)) and (a+h, f(a+h)). As h→0 the secant approaches the tangent, and its slope approaches the derivative.
Physical interpretation. In physics, if s(t) measures position at time t, then v(t)=s'(t) is velocity — the rate of change of position. Similarly acceleration is a(t)=v'(t)=s''(t), the rate of change of velocity. Understanding derivatives as rates gives direct modelling power for motion, growth, chemistry rates and many applied problems.
Existence and relationship with continuity. Differentiability implies continuity: if f is differentiable at a, then f is continuous at a. The converse is not true; functions may be continuous but not differentiable, for example at corners or cusps where left and right derivatives differ or blow up. Vertical tangents correspond to infinite slope and derivative may not exist in the finite sense.
Higher derivatives and notation. The second derivative f''(a) is derivative of f'(x) and measures how the slope itself changes: it describes concavity. Notation includes f'(x), f''(x), and Leibniz's form dy/dx emphasizing variables. Higher-order derivatives appear in series expansions and motion problems (jerk, snap).
Tangent and normal lines. If f is differentiable at a, tangent line equation is y − f(a) = f'(a)(x − a). The normal line at the same point is perpendicular to tangent and has slope −1/f'(a) when f'(a)≠0; its equation is y − f(a) = −1/f'(a) (x − a). Tangent and normal are used in geometry and applied optimisation.
Using the definition in practice. First-principle calculations are instructive for simple functions and to verify formulae, but for more complex functions use derivative rules. Nevertheless, always keep the limit definition in mind for conceptual clarity and for handling tricky definitions, piecewise behaviours and proofs.
- Find derivative of f(x)=x^2 at x=3 using first principle: f'(x)=2x so f'(3)=6.
- If s(t)=5t^2, velocity v(t)=s'(t)=10t and acceleration a(t)=s''(t)=10.
- Equation of tangent to y=x^3 at x=1: f'(x)=3x^2 ⇒ f'(1)=3; tangent: y−1=3(x−1).
- \[f'(a) = lim_{h→0} [f(a+h) − f(a)]/h\]
- Tangent line at a: y − f(a) = f'(a)(x − a)
- Normal line slope = −1/f'(a) (if f'(a)≠0)
Differentiation rules and standard derivatives
Why rules are useful. The definition of derivative is fundamental but applying limits each time is slow. Differentiation rules let us compute derivatives quickly and correctly by combining simpler results. These rules come from the limit definition and properties of limits.
Linearity and constants. Differentiation is linear: (c f(x))' = c f'(x) for constant c, and (f+g)' = f' + g'. The derivative of a constant function is zero. These simple rules let you separate complex expressions into sums and scale derivatives appropriately.
Power rule and polynomials. The power rule d/dx[x^n] = n x^{n−1} holds for any real n where the function makes sense; for polynomials it is the primary tool. Apply termwise to differentiate polynomials quickly.
Product and quotient rules. For product of two functions f and g: (f g)' = f' g + f g'. This follows from the limit of difference quotients. For quotient f/g with g≠0, (f/g)' = (f' g − f g')/g^2. These rules handle combinations of functions where multiplication or division occurs.
Standard derivatives. Know derivatives of key elementary functions: d/dx[e^{x}] = e^{x}; d/dx[a^{x}] = a^{x} ln a; d/dx[ln x] = 1/x for x>0. Trigonometric derivatives: d/dx[sin x]=cos x, d/dx[cos x]=−sin x, d/dx[tan x]=sec^2 x. These are building blocks for more complex expressions.
Chain rule for composites. For a composite y = f(g(x)), the chain rule gives dy/dx = f'(g(x))·g'(x). This is crucial: differentiate outer function at inner, then multiply by derivative of inner. Many exam problems combine algebraic, trig and exponential parts that require chain rule repeatedly.
Implicit differentiation. Not all functions are given explicitly as y=f(x). For an equation involving x and y, differentiate both sides with respect to x, treating y as y(x) and applying chain rule to terms involving y. Solve algebraically for dy/dx. This method handles circles, ellipses and other implicit curves.
Parametric derivatives. When x and y are given in terms of a parameter t, the derivative dy/dx = (dy/dt)/(dx/dt) provided dx/dt ≠ 0. This is derived by chain rule and is used for motion along a curve or parametric graphing.
Putting rules together. Complex derivatives often require product, chain and quotient rules combined. Work step by step, simplify where possible, and check by differentiating numerically or by alternative method for verification.
- Differentiate f(x)=x^3·sin x: use product rule to get f'(x)=3x^2 sin x + x^3 cos x.
- Differentiate y = (3x^2 + 1)^5: chain rule gives y' = 5(3x^2+1)^4 · 6x = 30x(3x^2+1)^4.
- Given x = cos t, y = sin t, find dy/dx: dy/dt = cos t, dx/dt = −sin t ⇒ dy/dx = −cot t.
- (f·g)' = f' g + f g'
- (f/g)' = (f' g − f g')/g^2
- \[d/dx[x^n] = n x^{n−1}\]
- \[d/dx[e^{x}] = e^{x}\]\[d/dx[ln x] = 1/x\]
- Chain rule: d/dx f(g(x)) = f'(g(x)) g'(x)
Higher order derivatives and Leibniz notation
Definition and notation. Higher order derivatives are repeated derivatives: the second derivative f''(x) = d/dx[f'(x)] = d^2y/dx^2 measures how the first derivative changes. The third derivative f^{(3)} is d/dx[f''(x)], and in general f^{(n)}(x) denotes the n-th derivative. Leibniz notation d^n y/dx^n records the order and emphasises the differentiated variable; it is especially helpful in differential equations and implicit differentiation.
Meaning and interpretation. The first derivative f' measures the instantaneous rate of change or slope. The second derivative f'' measures the rate of change of slope: it tells whether the slope is increasing or decreasing. If f''(x)>0 on an interval, the function is concave up and tangent lines lie below the curve; if f''(x)<0 it is concave down. Higher derivatives describe finer aspects of change: in physics, third derivative is called jerk (rate of change of acceleration) and appears in some motion models.
Calculating higher derivatives. For simple functions, higher derivatives follow a pattern: for y=e^{ax}, each differentiation multiplies by a so d^n y/dx^n = a^n e^{ax}. Polynomials eventually differentiate to zero after finite steps. For trigonometric functions, derivatives cycle: derivatives of sin x produce cos x, −sin x, −cos x, then return to sin x after four derivatives.
Using Leibniz notation. Leibniz form is convenient when differentiating implicitly. If an equation involves x and y, differentiating both sides with respect to x using d/dx leads to terms like dy/dx. For second derivative differentiate dy/dx with respect to x again; when differentiating expressions containing dy/dx treat it as a function of x and apply product or chain rules as needed.
Second derivative test and concavity. The second derivative helps classify stationary points: if f'(a)=0 and f''(a)>0 then f has a local minimum at a; if f''(a)<0 then it has a local maximum. If f''(a)=0 the test is inconclusive and one must examine higher derivatives or sign changes in f' to reach a conclusion.
Implicit higher derivatives. Implicit differentiation for higher derivatives requires differentiating the relation repeatedly and substituting previously found derivatives. For example, given x^2 + y^2 = r^2, dy/dx = −x/y; then d^2y/dx^2 can be found by differentiating dy/dx with respect to x and substituting dy/dx to eliminate y'. This process is algebraically heavier but systematic.
Applications. Higher derivatives are used in Taylor series (derivatives at a point appear as coefficients), differential equations of higher order, and advanced modelling in physics and engineering. Practice computing several orders builds fluency and pattern recognition.
- Find first and second derivatives of y = x^4: y' = 4x^3, y'' = 12x^2.
- For y given implicitly by x^3 + y^3 = 6xy, find dy/dx and d^2y/dx^2 by differentiating implicitly twice.
- Compute third derivative of y = e^{2x}: y' = 2e^{2x}, y'' = 4e^{2x}, y^{(3)} = 8e^{2x}.
- \[f^{(n)}(x) = d^n y/dx^n\]
- \[If y = e^{ax} then d^n y/dx^n = a^n e^{ax}\]
Rolle's theorem and the Mean Value Theorem
Rolle's theorem — statement and meaning. Rolle's theorem applies to a function f that is continuous on a closed interval [a,b], differentiable on the open interval (a,b), and satisfies f(a) = f(b). The theorem guarantees there exists at least one point c in (a,b) where the derivative f'(c) = 0. Geometrically, if the function begins and ends at the same height and is smooth in between, the curve must have a horizontal tangent somewhere.
Mean Value Theorem (MVT) — generalisation. The Mean Value Theorem relaxes the equal endpoint condition. If f is continuous on [a,b] and differentiable on (a,b), then there exists c in (a,b) such that f'(c) = [f(b) − f(a)]/(b − a). That is, the instantaneous rate of change at some interior point equals the average rate of change over the whole interval. The MVT is fundamental in analysis and has many consequences.
Why hypotheses matter. Continuity on the closed interval and differentiability on the open interval are essential. If continuity fails (a jump) or differentiability fails (a corner), conclusions may not hold. For example, the absolute value function on [−1,1] has f(−1)=f(1) but is not differentiable at 0, and there is no point with derivative zero in the classical sense.
Applications and consequences of MVT. One immediate consequence: if f'(x) = 0 for every x in an interval then f is constant on that interval. MVT is used to estimate function values, prove inequalities, and justify uniqueness results. For example it helps show that if two differentiable functions have equal derivatives on an interval then they differ by a constant. It also provides error bounds when approximating functions by linearisation: f(x) ≈ f(a) + f'(c)(x−a) for some c between a and x.
Using MVT in problem solving. To apply MVT, verify continuity on [a,b] and differentiability on (a,b), compute the average rate and solve f'(c) = that average. Problems often require finding the c explicitly when f' is invertible or showing existence without computing it. Rolle's theorem is used to show existence of stationary points between equal-valued points and is a tool in root-counting arguments for polynomials.
Exam style and examples. Typical ICSE problems ask to state the theorem and apply it to simple functions, or to use it to deduce properties like monotonicity or constancy. Constructing a function that meets or violates hypotheses is a common exercise. Understanding the geometric picture — secant line and tangent parallel to it — helps retain the theorem intuitively and apply it correctly in proofs.
- Apply Rolle's theorem to f(x)=sin x on [0,π] since f(0)=0=f(π), there exists c in (0,π) with f'(c)=cos c=0 so c=π/2.
- Use MVT on f(x)=x^2 on [1,3]: average rate=(9−1)/(2)=4 ⇒ f'(c)=2c=4 so c=2.
- Show that if f'(x)=0 for all x in (a,b) then f is constant by MVT: for any x1,x2 pick c with f'(c)=(f(x2)−f(x1))/(x2−x1)=0 ⇒ f(x2)=f(x1).
Maxima and minima of functions
Definitions and distinctions. A local (or relative) maximum of f at x=a occurs when f(a) ≥ f(x) for x in some small neighbourhood around a. A local minimum occurs when f(a) ≤ f(x) nearby. Global (absolute) maxima or minima compare values across the entire domain. Critical points — where f'(x)=0 or f' does not exist — are candidate points for local extrema, but must be tested.
First derivative test. The sign of the first derivative around a critical point tells us whether it is a maximum or minimum. If f' changes from positive to negative at a, the function increases then decreases, so a local maximum occurs. If f' changes from negative to positive, the function decreases then increases, giving a local minimum. If f' does not change sign, the point is not an extremum; it may be a point of inflection or a stationary plateau.
Second derivative test. When f'(a)=0 and f''(a) exists, the second derivative provides a fast test: f''(a)>0 implies a local minimum (concave up), f''(a)<0 implies a local maximum (concave down). If f''(a)=0 the test is inconclusive and one must use higher derivatives or return to the first derivative sign test.
Finding absolute extrema on closed intervals. For a continuous function on a closed interval [a,b], absolute maxima and minima exist. The method is: (1) find all critical points inside (a,b), (2) evaluate the function at these critical points and at the endpoints a and b, (3) the largest value is the absolute maximum and the smallest is the absolute minimum. Always check endpoints because they can supply the global extrema even if interior critical points exist.
Constrained optimisation (brief view). Many problems involve maximizing or minimizing subject to a constraint, e.g., maximise area given perimeter. Within ICSE scope, this is usually solved by substituting the constraint to reduce the problem to one variable and then differentiating. For more advanced methods, Lagrange multipliers are used but are beyond scope.
Practical steps and pitfalls. Always check domain restrictions, points where derivative does not exist, and endpoints. Sketching the function or using a sign-chart for f' helps. For rational functions check vertical asymptotes, since they can prevent global extrema or create unbounded ranges. For piecewise functions test boundary points where the formula changes.
Applications and examples. Extrema are crucial in physics (maximum range, minimum energy), economics (cost minimisation), and geometry (largest area or volume under constraints). Practice on polynomials, rationals and trigonometric functions builds intuition for behaviour and sign changes.
- Find maxima/minima of f(x)=x^3−3x: f'(x)=3x^2−3 => roots x=±1. f''(x)=6x, at x=1 f''=6>0 (min), at x=−1 f''=−6<0 (max).
- On [0,2], find absolute extrema of f(x)=x^2−4x+1: critical at x=2 where f'(2)=0; evaluate f(0)=1, f(2)=−3 ⇒ absolute minimum at x=2, maximum at x=0.
- Maximise area of rectangle with fixed perimeter 20: let sides x,y with 2(x+y)=20 ⇒ y=10−x. Area A=x(10−x)=10x−x^2. A' = 10−2x = 0 ⇒ x=5 gives max area 25.
- Critical points satisfy f'(x)=0 or f' undefined
- Second derivative test: f''(a)>0 ⇒ local min, f''(a)<0 ⇒ local max
Concavity and points of inflection
Meaning of concavity. Concavity describes the way a curve bends. If a curve lies above all its tangents on an interval, it is concave up (convex); if it lies below its tangents it is concave down. The second derivative f''(x) formalises this: f''(x)>0 indicates concave up, f''(x)<0 indicates concave down. Intuitively, concave up looks like a cup 2, concave down like an upside-down cup.
Points of inflection defined. A point of inflection is a point on the curve where the concavity changes sign, from up to down or down to up. At such a point the second derivative is often zero or undefined, but the key test is that f'' changes sign when moving through the point. Not every point where f''=0 is an inflection point; e.g., f(x)=x^4 has f''(0)=0 but does not change concavity there.
Finding inflection points. To locate possible inflection points compute f''(x) and find where f''(x)=0 or f'' undefined. Then check intervals on either side of each candidate to see if the sign of f'' changes. If it does, the point is an inflection. Check the function is defined there. For implicit curves or parametric forms compute d^2y/dx^2 appropriately and perform sign checks.
Relation to slope and first derivative. Concavity is about how the slope changes. If f' is increasing on an interval (f' rising as x increases), then f''>0 and function is concave up. If f' is decreasing, f''<0 and function is concave down. Therefore, examining the monotonicity of f' gives an alternative approach to understanding concavity when computing f'' is difficult.
Practical examples and common traps. For polynomials like cubic functions, an inflection typically occurs where f''=0 and sign changes, as in y=x^3 at the origin. For functions like y=x^4 there is no sign change despite f''(0)=0. For rational functions, vertical asymptotes may interrupt intervals and must be considered separately. Always test sign change; do not assume f''=0 implies an inflection.
Use in graph sketching and applications. Concavity information combined with increasing/decreasing tests gives a more accurate graph sketch: it tells how steepness changes and helps place inflection points which are important in economics (change of marginal returns), physics (change of acceleration), and engineering design. Knowing where curvature changes is important for approximations and for interpreting how functions behave beyond simply increasing or decreasing.
- For f(x)=x^3−3x, f''(x)=6x. Solve 6x=0 ⇒ x=0. Test sign: for x<0 f''<0, for x>0 f''>0 so x=0 is an inflection point.
- Find inflection points of y=sin x: y''=−sin x. Solve −sin x = 0 ⇒ x = nπ. Check sign change to confirm which are inflection points (they are at nπ where concavity changes).
- Consider y = x^4. y'' = 12x^2 ≥ 0 with equality at x=0, but sign does not change so x=0 is not an inflection point.
- Concave up ⇔ f''(x)>0, concave down ⇔ f''(x)<0
- Inflection point when f'' changes sign at that point
L'Hospital's rule and indeterminate forms
What are indeterminate forms? When taking limits, expressions like 0/0 or ∞/∞ are called indeterminate because they do not determine a unique limit. Other indeterminate forms include 0·∞, ∞−∞, 0^0, 1^∞ and ∞^0. Evaluating such limits requires algebraic manipulation, substitution, or special rules like L'Hospital's rule. The key is to transform the expression into a form where a standard limit or derivative can be applied.
L'Hospital's rule — statement and conditions. If f and g are differentiable functions near a (except possibly at a) and lim_{x→a} f(x) = lim_{x→a} g(x) = 0 or both ±∞, and g'(x) ≠ 0 near a, then lim_{x→a} f(x)/g(x) = lim_{x→a} f'(x)/g'(x) provided the latter limit exists (or is ±∞). The rule may be applied repeatedly if the derivative quotient still yields an indeterminate form. It is essential to check the hypotheses before applying the rule.
How to use the rule. First confirm the original limit is 0/0 or ∞/∞. Differentiate numerator and denominator separately, simplify, and re-evaluate the limit. If it still yields indeterminate form, repeat. For limits at infinity, the same idea applies: if both numerator and denominator tend to ±∞, replace by derivatives and evaluate limit of derivative quotient.
Other indeterminate forms — transformations. For forms like 0·∞, rewrite as a quotient: f(x)·g(x) = f(x)/(1/g(x)) and look for 0/0 or ∞/∞. For differences giving ∞−∞, get a common denominator or factor to convert to a quotient. For powers such as 0^0, 1^∞, or ∞^0, use logarithms: set y = u(x)^{v(x)} and consider ln y = v(x) ln u(x). Evaluate the limit of ln y using L'Hospital if required, and then exponentiate the result to obtain the limit of y.
Examples and caution. L'Hospital is powerful but sometimes unnecessary: algebraic simplification or known limits may be simpler. Also do not apply L'Hospital to expressions that are not in quotient form until they are converted to such. Errors often arise from differentiating incorrectly or ignoring domain conditions. When using repeated L'Hospital, ensure each application still meets the rule's hypotheses.
Practical advice for exams. Show clearly that the original form is indeterminate, perform differentiation step-by-step, and simplify intermediate expressions. If using logarithms for power limits show the transformation and the return to exponential form. These steps make the solution clear and acceptable in board exams.
- Compute lim_{x→0} (sin x − x)/x^3: both numerator and denominator →0; apply L'Hospital repeatedly to get limit = −1/6.
- Find lim_{x→∞} (ln x)/x: form ∞/∞, apply L'Hospital to get limit 0.
- Evaluate lim_{x→0+} x^x: write ln y = x ln x → 0·(−∞) rewrite as ln x/(1/x) and use L'Hospital to get 0 ⇒ y→1.
- If lim f = lim g = 0 or ±∞, lim f/g = lim f'/g' (if latter exists)
- \[For y(x)=u(x)^{v(x)}\]\[ln y = v(x) ln u(x) to transform indeterminate powers\]
Taylor and Maclaurin series (up to second/third order)
Idea and purpose of Taylor approximations. Many smooth functions can be approximated locally by polynomials using their derivatives at a point. A Taylor polynomial of degree n approximates the function near a point a by matching derivatives up to order n. This provides simple polynomial approximations that are easy to compute and useful for estimating values, evaluating limits, and understanding local behaviour.
General Taylor formula (finite form). The Taylor polynomial of degree n for f about a is P_n(x) = f(a) + f'(a)(x−a) + f''(a)(x−a)^2/2! + ... + f^{(n)}(a)(x−a)^n/n!. The remainder term R_n(x) measures the error: one common form is Lagrange's remainder R_n(x) = f^{(n+1)}(ξ)(x−a)^{n+1}/(n+1)! for some ξ between a and x. For small |x−a| and bounded higher derivative, R_n is small and the polynomial approximates f well.
Maclaurin series as special case. When a = 0 the Taylor series is called the Maclaurin series. Key Maclaurin expansions used frequently: e^{x} = 1 + x + x^2/2! + x^3/3! + ..., sin x = x − x^3/3! + x^5/5! − ..., cos x = 1 − x^2/2! + x^4/4! − .... Truncating these gives good approximations for small x; the number of correct terms depends on how many terms are included and the magnitude of x.
Low-order approximations and linearisation. The first-degree Taylor polynomial gives linear approximation (tangent line): f(x) ≈ f(a) + f'(a)(x−a). This linearisation is used to approximate function values close to a and to estimate small changes: Δf ≈ f'(a) Δx. Second- and third-order polynomials include curvature and improve accuracy.
Using series to evaluate limits and integrals. Series expansions help resolve indeterminate limits by replacing functions by their leading polynomial terms. For example use sin x ≈ x − x^3/6 to evaluate limits involving small angles. Maclaurin polynomials are also integrated termwise to approximate definite integrals when antiderivatives are complicated.
Practical classroom scope. ICSE/ISC expects students to derive and use Taylor or Maclaurin polynomials up to second or third order and to understand the remainder concept qualitatively. You should be able to expand simple functions around 0 or other points, use expansions for approximations and limits, and estimate error where asked. Memorise common series and practise deriving a few low-order polynomials by differentiating at the expansion point.
- Find Maclaurin polynomial of degree 3 for e^{x}: P_3(x)=1+x+x^2/2 + x^3/6.
- Approximate sin 0.2 using sin x ≈ x − x^3/6 ⇒ sin 0.2 ≈ 0.2 − (0.008)/6 = 0.2 − 0.001333 = 0.198667.
- Use Taylor about a=1 to approximate ln x near x=1: ln x ≈ (x−1) − (x−1)^2/2 for small |x−1|.
- \[Taylor series at a: f(a)+f'(a)(x−a)+f''(a)(x−a)^2/2!+...+f^{(n)}(a)(x−a)^n/n!\]
- \[Maclaurin for e^{x}\]\[sin x\]\[cos x as given in explanation\]
Antiderivatives and indefinite integration
Antiderivative concept. An antiderivative F(x) of f(x) is a function whose derivative is f(x); i.e., F'(x)=f(x). The indefinite integral ∫ f(x) dx denotes the family of all antiderivatives and includes an arbitrary constant C because differentiation loses constant information. Finding antiderivatives is the reverse process of differentiation and forms the first part of integration techniques.
Basic integration rules and linearity. Integration is linear: ∫ [α f(x) + β g(x)] dx = α ∫ f(x) dx + β ∫ g(x) dx for constants α,β. The power rule, reverse of the differentiation rule, gives ∫ x^n dx = x^{n+1}/(n+1) + C for n≠−1. Remember special cases: ∫ 1/x dx = ln|x| + C, ∫ e^{x} dx = e^{x} + C, ∫ sin x dx = −cos x + C, ∫ cos x dx = sin x + C. These are the basic building blocks.
Substitution method. Substitution (change of variable) is used when the integrand is a composite function and its inner derivative appears. Let u = g(x) and du = g'(x) dx; then ∫ f(g(x)) g'(x) dx becomes ∫ f(u) du. Choosing an appropriate substitution simplifies the integral to a known form. For definite integrals change the limits to u-values to evaluate directly without back-substitution.
Integration by parts. Based on reversing the product rule, ∫ u dv = u v − ∫ v du. Choose u as a factor that simplifies on differentiation (logarithmic, inverse trigonometric or algebraic) and dv as a factor whose integral v is manageable. Repeated integration by parts is commonly used for integrals involving x^n e^{x} or x^n sin x and tabular integration offers a structured approach for repeated applications.
Partial fractions for rational functions. When integrating P(x)/Q(x) with degrees such that numerator degree < denominator degree and Q factors over reals, decompose the fraction into simpler partial fractions. Integrate each simpler term using logarithms for linear denominators and arctan for irreducible quadratics after completing the square.
Verification and constant of integration. After finding an antiderivative, differentiate it to verify the original integrand. Always include +C for indefinite integrals and determine C using initial conditions when solving differential equations or initial-value problems. Practice combining techniques: substitution inside parts, parts after partial fractions and other mixes are common in ICSE/ISC questions.
- Find ∫ (3x^2 − 2x + 1) dx = x^3 − x^2 + x + C.
- Integrate ∫ e^{2x} dx: let u=2x, du=2dx ⇒ ∫ e^{2x} dx = e^{2x}/2 + C.
- Compute ∫ x cos x dx by parts: take u=x, dv=cos x dx ⇒ du=dx, v=sin x ⇒ x sin x − ∫ sin x dx = x sin x + cos x + C.
- \[∫ x^n dx = x^{n+1}/(n+1) + C (n≠−1)\]
- ∫ u dv = uv − ∫ v du
- Substitution: ∫ f(g(x)) g'(x) dx = ∫ f(u) du
The definite integral and Fundamental Theorem of Calculus
Definite integral as limit of Riemann sums. The definite integral ∫_{a}^{b} f(x) dx is defined as the limit of Riemann sums that approximate the signed area under the curve y=f(x) from x=a to x=b. Partition [a,b] into n subintervals, pick sample points x_i^*, form sum Σ f(x_i^*) Δx_i and take the limit as the maximum Δx_i tends to 0. This formalism captures accumulation of infinitesimal contributions and makes integration rigorous.
Properties of definite integrals. Integrals are linear: ∫_{a}^{b} [α f + β g] = α ∫_{a}^{b} f + β ∫_{a}^{b} g. They are additive over intervals: ∫_{a}^{c} f + ∫_{c}^{b} f = ∫_{a}^{b} f. Reversing limits changes sign: ∫_{b}^{a} f = −∫_{a}^{b} f. If f(x) ≥ 0 on [a,b] then ∫_{a}^{b} f(x) dx ≥ 0. These properties aid manipulation and splitting of integrals in calculation and proofs.
Fundamental Theorem of Calculus (two parts). Part 1: If F(x)=∫_{a}^{x} f(t) dt and f is continuous on [a,b], then F is differentiable and F'(x)=f(x). This shows integration accumulates the integrand and differentiation recovers it. Part 2: If f is continuous on [a,b] and F is any antiderivative of f on [a,b], then ∫_{a}^{b} f(x) dx = F(b) − F(a). This connects antiderivatives to definite integrals and is the main computational tool allowing evaluation of integrals by finding primitives instead of limits of sums.
Signed area and regions crossing the axis. The definite integral gives signed area: regions where f≥0 contribute positive area, where f<0 contribute negative. To find the total geometric area between a curve and the axis, integrate absolute value or split at zeros where sign changes and add absolute values of integrals on subintervals.
Evaluating definite integrals practically. To compute ∫_{a}^{b} f(x) dx use FTC part 2: find an antiderivative F, then compute F(b)−F(a). For integrals requiring substitution, change the limits to u-values or revert after integration. For improper integrals replace infinite limits or singular points by limits and test for convergence.
Applications and interpretation. Definite integrals compute areas, displacements, accumulated quantities like mass or charge, and probabilities in continuous distributions. Understanding both the limit definition and FTC gives both conceptual and practical mastery: the definition explains why integration produces area, FTC allows efficient calculation in exam problems.
- Compute ∫_{0}^{1} 3x^2 dx: antiderivative x^3 gives 1^3 − 0 = 1.
- If F(x)=∫_{2}^{x} (1/t) dt then F'(x)=1/x by FTC part 1.
- Evaluate ∫_{0}^{π} sin x dx = [−cos x]_{0}^{π} = (−cos π)−(−cos 0)=2.
- \[∫_{a}^{b} f(x) dx = F(b) − F(a) where F' = f\]
- \[If F(x)=∫_{a}^{x} f(t) dt then F'(x)=f(x)\]
Techniques of integration: substitution and parts
Substitution: change of variable explained. Substitution transforms an integral into a simpler one by setting u=g(x) so that du = g'(x) dx. The idea is to recognise an inner function whose derivative multiplies the rest of the integrand. Once substituted, the integral becomes ∫ f(u) du which is often straightforward. For definite integrals, convert the limits from x-values to corresponding u-values to evaluate directly without back-substitution. Substitution is the counterpart of the chain rule in differentiation.
Choosing substitutions. Good substitution choices often make a composite expression simpler: if integrand contains (ax+b)^n times its derivative, set u=ax+b. For trigonometric integrals, substitutions like u=sin x or u=cos x often reduce the power. When integrand contains sqrt expressions, substitution like u = sqrt(...) or trigonometric substitution can rationalise the integrand.
Integration by parts — reversing product rule. Integration by parts is based on the product rule: (u v)' = u' v + u v'. Rearranged and integrated, ∫ u dv = u v − ∫ v du. The choice of u and dv is strategic: pick u to be a function that gets simpler when differentiated (logarithmic, inverse trig, algebraic) and pick dv to be easily integrable. The LIATE (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential) rule helps choose u in many cases.
Repeated parts and tabular integration. When repeated application of parts is needed (e.g., ∫ x^n e^{x} dx), tabular integration provides a structured shortcut: list successive derivatives of u and successive integrals of dv, then combine with alternating signs. This saves time and reduces algebraic mistakes in exams.
Combining methods and special cases. Many integrals require substitution followed by parts or vice versa. For instance, ∫ x e^{x^2} dx uses substitution u = x^2 (so du = 2x dx) and becomes straightforward. Definite integrals should adjust limits or transform back to x after integrating. Always verify results by differentiating the antiderivative and checking constant of integration for indefinite integrals.
Common patterns and practice guidance. Memorise standard substitution patterns and practise parts on polynomials times exponentials or trigonometric functions. Learn to spot when integration by parts reduces the power of a polynomial factor. Show steps clearly in exams: state substitution, change limits if used, and simplify intermediate expressions. This clarity prevents small algebraic errors and earns full credit.
- Compute ∫ x e^{x} dx by parts: u=x, dv=e^{x} dx ⇒ du=dx, v=e^{x} ⇒ ∫ x e^{x} dx = x e^{x} − ∫ e^{x} dx = x e^{x} − e^{x} + C.
- Integrate ∫ x(1+x^2)^4 dx by substitution: u = 1+x^2, du = 2x dx ⇒ integral = (1/2) ∫ u^4 du = (1/10) u^5 + C.
- Use tabular method for ∫ x^2 e^{x} dx to get e^{x}(x^2 − 2x + 2) + C.
- Substitution: ∫ f(g(x)) g'(x) dx = ∫ f(u) du
- Integration by parts: ∫ u dv = u v − ∫ v du
Integration of rational functions and partial fractions
Rational functions and proper fractions. A rational function is a quotient P(x)/Q(x) of polynomials. For integration we prefer proper fractions where degree of numerator < degree of denominator. If the numerator degree ≥ denominator degree, perform polynomial long division to separate out a polynomial part and a proper fraction, then integrate each part separately.
Partial fraction decomposition. Once proper, factor the denominator Q(x) over real numbers as a product of linear factors (x−r) and irreducible quadratic factors. Set up an expression where the proper fraction equals a sum of simpler terms: for distinct linear factors use A/(x−r); for repeated linear factors use A_1/(x−r) + A_2/(x−r)^2 + ...; for irreducible quadratics use (Bx+C)/(x^2 + px + q). Solve for constants by clearing denominators and equating coefficients or by substituting convenient x-values when possible.
Integrating decomposed terms. Integrals of A/(x−a) give A ln|x−a|. Terms of the form (Bx+C)/(x^2+px+q) can be split: the derivative of denominator provides a logarithmic part and the remaining constant part integrates to an arctan after completing the square. For irreducible quadratics, write denominator as (x + m)^2 + n^2 and use substitution u = x+m to obtain (1/n) arctan(u/n) forms. For repeated linear factors integrate powers (x−a)^{−n} using the power rule for n ≠ 1.
Procedure and algebraic care. Steps: (1) If needed perform polynomial division. (2) Factor denominator. (3) Propose partial fraction form with unknown constants. (4) Solve for constants. (5) Integrate each resulting simple term. Algebraic accuracy is essential; errors in solving constants lead to incorrect integrals. Show workings clearly in exams.
Special cases and examples. Some integrals require completing the square before decomposition, or may include linear substitutions before partial fractions make sense. Practice standard examples like ∫ (2x+3)/(x^2+x−2) dx and ∫ 1/(x^2+4x+5) dx to build technique. Knowing standard integrals ∫ dx/(x−a) = ln|x−a| and ∫ dx/(x^2 + a^2) = (1/a) arctan(x/a) is crucial.
Applications in definite integrals. For definite integrals apply decomposition and evaluate antiderivatives at limits. For improper integrals of rational functions, test convergence with degree comparisons and p-integral criteria when integrating over infinite ranges.
- Integrate ∫ (2x+3)/(x^2+x−2) dx. Factor denominator (x+2)(x−1) and write partial fractions A/(x+2)+B/(x−1), solve for A,B then integrate to get logarithms.
- Find ∫ 1/(x^2+4x+5) dx by completing square: denominator (x+2)^2+1 ⇒ substitute u=x+2 and get arctan form giving arctan(x+2).
- Compute ∫ (x^2+1)/(x^3 − x) dx by partial fractions after factoring denominator x(x−1)(x+1).
- ∫ dx/(x−a) = ln|x−a| + C
- ∫ dx/(x^2 + a^2) = (1/a) arctan(x/a) + C
Improper integrals and convergence
What makes an integral improper? An integral is improper when the interval of integration is infinite (e.g., ∫_{a}^{∞} f(x) dx) or when the integrand becomes unbounded at some point inside the interval (e.g., ∫_{0}^{1} 1/√x dx where integrand has a singularity at 0). Such integrals are defined as limits of proper integrals and must be tested for convergence (finite value) or divergence (infinite or undefined).
Evaluation by limits. For an improper integral over an infinite interval, compute lim_{R→∞} ∫_{a}^{R} f(x) dx. For an integrand with singularity at endpoint b, evaluate lim_{ε→0+} ∫_{a}^{b−ε} f(x) dx (or split integrals around interior singularities). If the limit exists and is finite, the integral converges; otherwise it diverges. For integrals with singularities at both ends split into two limits and check each part separately.
Comparison tests. Comparison tests are practical: if 0 ≤ f(x) ≤ g(x) for x large and ∫ g converges, then ∫ f converges. Conversely, if f(x) ≥ g(x) ≥ 0 and ∫ g diverges, then ∫ f diverges. This allows using simple known integrals to judge more complicated ones. Typical comparison targets are p-integrals ∫_{1}^{∞} 1/x^{p} dx which converge if p>1 and diverge if p≤1.
p-integral rules and near-zero behaviour. The p-integral test for behaviour near a singularity says ∫_{0}^{1} 1/x^{p} dx converges if p<1 and diverges if p≥1. For infinite intervals, ∫_{1}^{∞} 1/x^{p} dx converges if p>1 and diverges otherwise. Use these as benchmarks for comparing decay rates of functions as x→∞ or blow-up rates as x→0.
Improper integrals with oscillation. Some integrals oscillate yet converge conditionally, for example ∫_{0}^{∞} sin x/x dx converges (the Dirichlet integral). Testing such integrals sometimes requires special tests (Dirichlet, Abel) beyond simple comparison. For ICSE level, focus on comparison, p-tests and limit definitions; note when special techniques are needed.
Applications and examples. Improper integrals appear in probability, physics (total work, energy over infinite domains), and in series-integral tests. Typical exam tasks ask to test convergence or compute convergent improper integrals using appropriate limits or comparisons. Always justify the limit process and state whether the integral converges or diverges.
- Determine ∫_{1}^{∞} 1/x^2 dx: evaluate limit ∫_{1}^{R} x^{−2} dx = 1 − 1/R ⇒ limit 1, so converges.
- Check ∫_{0}^{1} 1/√x dx: integral ∫_{ε}^{1} x^{−1/2} dx = 2(1 − √ε) ⇒ limit 2, so converges.
- Test ∫_{1}^{∞} 1/x dx by p-test with p=1 ⇒ diverges.
- \[p-integral: ∫_{1}^{∞} 1/x^{p} dx converges ⇔ p>1\]
- \[∫_{0}^{1} 1/x^{p} dx converges ⇔ p<1\]
Applications of integration: area between curves and volumes
Area between two curves. When two continuous functions f and g satisfy f(x) ≥ g(x) on [a,b], the area of the planar region bounded by their graphs and the vertical lines x=a and x=b is A = ∫_{a}^{b} [f(x) − g(x)] dx. To apply this method first sketch the curves or find intersection points where f(x)=g(x) to determine the correct limits and the interval on which one function is above the other. If the ordering changes, split the integral at intersection points and sum absolute areas on each subinterval.
Computational steps for area. Steps: (1) Find intersection points by solving f(x)=g(x). (2) Determine which function is above the other on each subinterval. (3) Set up the integral of top minus bottom and evaluate. For regions bounded by horizontal lines or when functions are better expressed as x=g(y), integrate with respect to y and use right minus left instead.
Volumes by revolution — disc and washer methods. Revolve a region around an axis to create a solid. If slices perpendicular to the axis of revolution produce circular cross-sections, use the disc method: V = π ∫_{a}^{b} [R(x)]^2 dx where R(x) is the radius from axis to curve. If the region has a hole, use the washer method: V = π ∫_{a}^{b} [R(x)^2 − r(x)^2] dx where R and r are outer and inner radii respectively.
Shell method. The shell method is useful when discs are awkward. For rotation around a vertical axis and vertical slices, a cylindrical shell of radius x and height h(x) has volume 2π x h(x) Δx; integrating gives V = 2π ∫_{a}^{b} x h(x) dx. Shells are often easier when rotating regions described as functions of x about vertical axes without rewriting in terms of y.
Choosing method and examples. Decide between disc/washer and shell by inspecting whether slices perpendicular or parallel to the axis produce simpler expressions. For example, revolving y = √x on [0,1] about the x-axis uses discs easily: V = π ∫_{0}^{1} (√x)^2 dx = π/2. For revolving the same region about the y-axis the shell method is simpler.
Applications and problem-solving tips. Integration-based area and volume calculations appear frequently in exams. Sketch the region carefully, label radii/height, and check limits. For volumes, ensure radii are measured as distances (absolute values) and that the integrand is non-negative for area. Use algebraic simplification to make antiderivatives manageable and explain steps clearly in written solutions.
- Find area between y=x and y=x^2 from 0 to 1: ∫_{0}^{1} (x − x^2) dx = 1/6.
- Volume of solid formed by revolving y = sqrt(x) from 0 to 1 around x-axis: π ∫_{0}^{1} (√x)^2 dx = π/2.
- Using shell method, volume formed by revolving region between y=x^2 and y=0 from x=0 to 1 about y-axis: 2π ∫_{0}^{1} x(x^2) dx = 2π ∫_{0}^{1} x^3 dx = π/2.
- \[Area between curves: ∫_{a}^{b} [f(x) − g(x)] dx\]
- \[Disc/washer: V = π ∫_{a}^{b} [R(x)^2 − r(x)^2] dx\]
- \[Shell method: V = 2π ∫_{a}^{b} (radius)(height) dx\]
Reduction formulae
What are reduction formulae? Reduction formulae provide a recurrence relation that reduces an integral involving a parameter n to another integral with a smaller parameter, typically n−1 or n−2. They are particularly helpful when dealing with families of integrals such as ∫ sin^n x dx, ∫ cos^n x dx or ∫ x^n e^{x} dx where direct integration for general n is difficult. By using integration by parts or trigonometric identities, one derives a formula linking I_n to I_{n−1} or I_{n−2} and then uses base cases to compute values.
Deriving a typical reduction formula. Consider I_n = ∫ sin^n x dx. Use integration by parts with u = sin^{n−1} x and dv = sin x dx; then du = (n−1) sin^{n−2} x cos x dx and v = −cos x, leading after algebra to a relation expressing I_n in terms of I_{n−2}. The algebraic steps often require rearranging and isolating I_n. For integrals with polynomial times exponential, use parts with u = x^n and dv = e^{x} dx to derive J_n = x^n e^{x} − n J_{n−1} where J_n = ∫ x^n e^{x} dx.
Using reduction formulae. Once a recurrence is derived, compute integrals for specific n by repeatedly applying the relation until reaching a base case that you can integrate directly. For example, to compute ∫ x^3 e^{x} dx apply the recurrence thrice until J_0 = ∫ e^{x} dx is obtained. For definite integrals from 0 to π/2 involving sin^n x, reduction formulas give exact values and are used to compute integrals for even and odd n separately.
Common reduction formulas. Examples include I_n = ∫ sin^n x dx = −(sin^{n−1} x cos x)/n + (n−1)/n I_{n−2}, and J_n = ∫ x^n e^{x} dx = x^n e^{x} − n J_{n−1}. For definite integrals like ∫_{0}^{π/2} sin^n x dx, reduction yields I_n = (n−1)/n · I_{n−2} leading to closed forms for even and odd n.
Practical advice and exam tips. When deriving reduction relations show each integration by parts step clearly and solve algebraically for the integral of interest. Keep track of signs and constants carefully. Use initial values such as I_0 and I_1 to start the recurrence. Reduction formula questions often test algebraic manipulation as much as integration technique, so neat presentation and accurate algebra help secure full marks.
- Derive I_n = ∫ sin^n x dx to show I_n = −(sin^{n−1} x cos x)/n + (n−1)/n I_{n−2}.
- Use J_n = ∫ x^n e^{x} dx to find J_2 by recurrence: J_2 = x^2 e^{x} − 2 J_1, J_1 = x e^{x} − e^{x} + C.
- Apply reduction to definite integral ∫_{0}^{π/2} sin^n x dx to relate I_n to I_{n−2} and evaluate for even n values.
- \[I_n = ∫ sin^n x dx = −(sin^{n−1} x cos x)/n + (n−1)/n I_{n−2}\]
- \[J_n = ∫ x^n e^{x} dx = x^n e^{x} − n J_{n−1}\]
First-order differential equations: separable and linear
Meaning and goals. A differential equation relates a function and its derivatives. First-order equations involve dy/dx (or dy/dt) only. Solving means finding the general solution family y(x) and, when an initial condition is given, a particular solution. First-order models are used for growth and decay, cooling, mixing and many physical problems.
Separable equations and method. A differential equation is separable if it can be written as dy/dx = g(x) h(y), allowing separation: dy/h(y) = g(x) dx. Integrate both sides to get ∫ dy/h(y) = ∫ g(x) dx + C and then solve algebraically for y if possible. This method gives solutions like exponential growth: dy/dt = k y ⇒ dy/y = k dt ⇒ ln|y| = kt + C ⇒ y = C' e^{kt}.
Linear first-order equations and integrating factor. A linear first-order equation has the form dy/dx + P(x) y = Q(x). Multiply by the integrating factor μ(x) = e^{∫ P(x) dx} which makes the left-hand side the derivative of μ(x) y: d/dx[μ(x) y] = μ(x) Q(x). Integrate both sides: μ(x) y = ∫ μ(x) Q(x) dx + C and divide by μ(x) to get the general solution y = [∫ μ Q dx + C]/μ. This method handles non-homogeneous terms Q(x) that are polynomials, trigonometric or exponential functions.
Initial value problems. Use the initial condition y(x_0) = y_0 to determine the constant C in the general solution. This yields a unique particular solution under usual continuity conditions. Check the solution by differentiating and substituting back into the original differential equation to verify correctness.
Physical examples and interpretation. Exponential growth and decay: dy/dt = ky leads to y = y_0 e^{kt}. Newton's law of cooling dy/dt = −k(y − T_{env}) leads to solution y − T_{env} = C e^{−kt}, showing exponential approach to ambient temperature. Mixing problems lead to linear equations where Q(x) depends on inflow and outflow rates and concentrations; integrating factor solves such problems when arranged in standard form.
Practical solving strategies. Recognise the type quickly: if variables separate, use separation; else try to write in linear form and use integrating factor. Simplify algebra early, and keep constants organised. For equations reducible by substitution (e.g., homogeneous form or Bernoulli type beyond basic scope), transform into separable or linear types when possible. Provide full working and check the final solution with the initial condition in exam answers.
- Solve dy/dx = 3y: separable ⇒ dy/y = 3 dx ⇒ ln|y| = 3x + C ⇒ y = C' e^{3x}.
- Solve dy/dx + y = x: integrating factor μ = e^{∫1 dx} = e^{x}. Then d/dx(e^{x} y) = x e^{x}. Integrate: e^{x} y = ∫ x e^{x} dx = e^{x}(x−1)+C ⇒ y = x−1 + Ce^{-x}.
- Newton cooling: dy/dt = −k(y − T_0) ⇒ y − T_0 = Ce^{−kt}.
- Separable form: ∫ dy/h(y) = ∫ g(x) dx + C
- \[Linear integrating factor: μ(x) = e^{∫ P(x) dx}\]\[then y = [∫ μ Q dx + C]/μ\]
Key Concepts
- Limit
- The value that a function approaches as the input approaches a point.
- Continuity
- A function is continuous at a point if its limit there equals its function value.
- Derivative
- The instantaneous rate of change or slope of the tangent to a curve at a point.
- Differentiability
- A function is differentiable at a point if its derivative (limit of difference quotient) exists there.
- Chain rule
- A rule to differentiate composite functions: (f∘g)' = f'(g(x))·g'(x).
- Rolle's theorem
- If f is continuous on [a,b], differentiable on (a,b) and f(a)=f(b), then some c in (a,b) has f'(c)=0.
- Mean Value Theorem
- There exists c in (a,b) with f'(c) equal to average rate [f(b)−f(a)]/(b−a) under continuity and differentiability.
- Inflection point
- A point where the function changes concavity from up to down or down to up.
- Indeterminate form
- A limit expression like 0/0 or ∞/∞ whose value cannot be determined without further work.
- L'Hospital's rule
- A method to evaluate limits of quotients in 0/0 or ∞/∞ form by differentiating numerator and denominator.
- Antiderivative
- A function whose derivative is the given function; represents indefinite integral plus constant.
- Definite integral
- The limit of Riemann sums representing signed area under a curve between two limits.
- Fundamental Theorem of Calculus
- Connects differentiation and integration: derivative of integral equals integrand and integral equals difference of antiderivative values.
- Partial fractions
- A method to decompose rational functions into simpler fractions for integration.
- Improper integral
- An integral where interval is infinite or integrand unbounded, evaluated as a limit.
- Integrating factor
- A function e^{∫P(x) dx} used to solve linear first-order differential equations.
- Taylor polynomial
- A finite-degree polynomial approximation to a function derived from its derivatives at a point.
- Rate of change
- How a quantity changes with respect to another, measured by derivative.
Practice Questions
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Find lim_{x→0} (sin x)/x. / lim_{x→0} (sin x)/x क्या है?
Show answer
Answer: lim_{x→0} (sin x)/x = 1. / उत्तर: lim_{x→0} (sin x)/x = 1.
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State and apply the Mean Value Theorem to f(x)=x^2 on [1,3]. / Mean Value Theorem को बयान कीजिए और f(x)=x^2 पर [1,3] पर लागू कीजिए।
Show answer
Answer: MVT: If f is continuous on [a,b] and differentiable on (a,b) then exists c with f'(c) = [f(b)−f(a)]/(b−a). For f(x)=x^2 on [1,3], average rate = (9−1)/(2)=4 so f'(c)=2c=4 ⇒ c=2. / उत्तर: MVT के अनुसार ऐसा c मौजूद है; यहाँ c=2 क्योंकि f'(x)=2x और 2·2=4.
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Differentiate y = ln(x^2 + 1). / y = ln(x^2 + 1) का अवकलन कीजिए।
Show answer
Answer: dy/dx = (2x)/(x^2 + 1). / उत्तर: dy/dx = 2x/(x^2 + 1).
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Find the equation of tangent to y = x^3 at x=1. / y = x^3 पर x=1 में स्पर्श रेखा का समीकरण求 करें।
Show answer
Answer: f'(x)=3x^2 so f'(1)=3. Point (1,1). Tangent: y−1 = 3(x−1) or y = 3x −2. / उत्तर: स्पर्श रेखा y−1 = 3(x−1) अर्थात y = 3x − 2.
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Evaluate ∫_{0}^{1} (3x^2) dx. / ∫_{0}^{1} (3x^2) dx का मान बताइए।
Show answer
Answer: Antiderivative x^3; value = 1^3 − 0 = 1. So integral = 1. / उत्तर: ∫ = [x^3]_{0}^{1} = 1.
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Using substitution evaluate ∫ x(1+x^2)^4 dx. / उपनिवेश विधि का प्रयोग कर ∫ x(1+x^2)^4 dx निकालिए।
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Answer: Let u=1+x^2 ⇒ du=2x dx ⇒ integral = (1/2) ∫ u^4 du = (1/10) u^5 + C = (1/10)(1+x^2)^5 + C. / उत्तर: u=1+x^2 से ∫ = (1/10)(1+x^2)^5 + C.
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Determine whether ∫_{1}^{∞} 1/x dx converges. / ∫_{1}^{∞} 1/x dx समाकलित होता है या नहीं?
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Answer: This is p-integral with p=1; it diverges (harmonic). ∫_{1}^{R} 1/x dx = ln R → ∞ as R→∞. / उत्तर: यह विसंमित है; मान ln R → ∞ अतः समाकलित नहीं होता।
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Solve the differential equation dy/dx = 3y with y(0)=2. / dy/dx = 3y और y(0)=2 को हल कीजिए।
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Answer: Separable: dy/y = 3 dx ⇒ ln|y| = 3x + C. Using y(0)=2 ⇒ ln 2 = C. So y = 2 e^{3x}. / उत्तर: y = 2 e^{3x}.
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Find ∫ (2x+3)/(x^2 + x − 2) dx by partial fractions. / आंशिक भिन्नों द्वारा ∫ (2x+3)/(x^2 + x − 2) dx निकालिए।
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Answer: Factor denominator: (x+2)(x−1). Write (2x+3)/(...) = A/(x−1)+B/(x+2). Solve: 2x+3 = A(x+2)+B(x−1). Setting x=1 gives 5 = A·3 ⇒ A=5/3. x=−2 gives −1 = B(−3) ⇒ B=1/3. Integral = (5/3) ln|x−1| + (1/3) ln|x+2| + C. / उत्तर: A=5/3, B=1/3, अतः ∫ = (5/3) ln|x−1| + (1/3) ln|x+2| + C.
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Find the area enclosed between y=x and y=x^2. / y=x और y=x^2 के बीच बंद क्षेत्र का क्षेत्रफल ज्ञात कीजिए।
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Answer: Intersection at x=0 and x=1. Area = ∫_{0}^{1} (x − x^2) dx = [x^2/2 − x^3/3]_{0}^{1} = 1/2 − 1/3 = 1/6. / उत्तर: क्षेत्रफल = 1/6.
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Use L'Hospital's rule to evaluate lim_{x→∞} (ln x)/x. / L'Hospital नियम द्वारा lim_{x→∞} (ln x)/x निकालिए।
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Answer: Form ∞/∞ so differentiate numerator and denominator: limit = lim_{x→∞} (1/x)/1 = lim_{x→∞} 1/x = 0. / उत्तर: सीमा 0 है।
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