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Class 12 Mathematics Chapter 7 of 13

Chapter 7 — Integrals

Overview

Chapter: Integrals (Class 12, NCERT) — This chapter develops the concept of integration as the reverse process of differentiation and as a limit of sums. It introduces indefinite integrals (antiderivatives) and standard integration formulae, methods of integration (substitution, integration by parts, partial fractions and trigonometric integrals), and definite integrals defined via Riemann sums. The Fundamental Theorem of Calculus linking differentiation and integration is presented and used to evaluate definite integrals. The chapter concludes with geometric applications, chiefly the computation of areas bounded by curves. Importance: integration is a core tool for solving problems in mathematics, physics and engineering — for finding areas, accumulated quantities and solving differential equations. What the student will learn: compute antiderivatives, apply a range of techniques to find integrals of algebraic, exponential, trigonometric and rational functions, evaluate definite integrals using the Fundamental Theorem, manipulate properties of integrals, and calculate areas under and between curves.

Learning Objectives

  • Define indefinite integral and antiderivative; state basic integration rules (linearity, power rule, constants).
  • Explain the definite integral as a limit of Riemann sums and state its key properties (additivity, linearity, positivity).
  • State and apply the Fundamental Theorem of Calculus to evaluate definite integrals by antiderivatives.
  • Evaluate indefinite integrals of algebraic, exponential, logarithmic, trigonometric and inverse trigonometric functions.
  • Apply the substitution (change of variable) method to compute indefinite and definite integrals.
  • Apply integration by parts to evaluate integrals of products, including those with logarithmic and inverse trigonometric factors.
  • Integrate rational functions by decomposing into partial fractions and evaluate the resulting integrals.
  • Use symmetry and other properties (even/odd functions, periodicity) to simplify and evaluate definite integrals.

Topics in this chapter

8 topics · tap a topic title to jump straight to it.

⚖️1

Introduction to Integration

What is Integration?

Integration is a fundamental operation in calculus that can be thought of in two closely related ways:

  • Inverse of differentiation: If F'(x) = f(x), then F(x) is called an antiderivative or an indefinite integral of f(x). We write F(x) = ∫ f(x) dx + C (C is an arbitrary constant).
  • Limit of sums / area interpretation: A definite integral ∫ab f(x) dx gives the signed area between the curve y = f(x) and the x-axis from x = a to x = b. Formally it is the limit of Riemann sums: ∫ab f(x) dx = limn→∞ Σ f(xi*) Δx.

Indefinite vs Definite Integrals

  • Indefinite integral: ∫ f(x) dx = F(x) + C, represents a family of antiderivatives.
  • Definite integral: ∫ab f(x) dx = F(b) − F(a) (when F is an antiderivative of f). This gives a number (signed area).

Fundamental Theorem of Calculus (FTC)

  • Part 1 (Connection): If F(x) = ∫ax f(t) dt and f is continuous, then F'(x) = f(x). So integration followed by differentiation returns the original function.
  • Part 2 (Evaluation): If F is any antiderivative of f on [a,b], then ∫ab f(x) dx = F(b) − F(a). This is the main tool to compute definite integrals.

When is a function integrable?

A bounded function that is continuous almost everywhere (or piecewise continuous) on [a,b] is Riemann integrable. For CBSE Class 12, typical functions considered are continuous or piecewise continuous.

Basic idea of computation

To compute indefinite integrals you use standard formulas (power rule, basic trigonometric integrals, exponential, logarithmic) and techniques (substitution, integration by parts). For definite integrals, you find an antiderivative and apply the FTC; alternatively, evaluate limits of sums for area problems.

📌 Examples
  • 1) Indefinite integral (power rule): ∫ x^3 dx = x^4/4 + C. Explanation: Increase power by 1 and divide by new power.
  • 2) Simple linear function: ∫ (2x + 3) dx = x^2 + 3x + C.
  • 3) Trigonometric integral: ∫ sin x dx = -cos x + C.
  • 4) Substitution example: ∫ 2x cos(x^2) dx. Let u = x^2 ⇒ du = 2x dx, so integral = ∫ cos u du = sin u + C = sin(x^2) + C.
  • 5) Integration by parts example: ∫ x e^x dx. Let u = x (du = dx), dv = e^x dx (v = e^x). Then ∫ x e^x dx = x e^x - ∫ e^x dx = x e^x - e^x + C = e^x(x - 1) + C.
  • 6) Definite integral (area): Find area under y = x^2 from x = 0 to x = 2. Compute ∫_0^2 x^2 dx = [x^3/3]_0^2 = (8/3) - 0 = 8/3 (square units).
🧮 Formulas
  1. Indefinite integral and constant: ∫ f(x) dx = F(x) + C, where F'(x) = f(x).
  2. Power rule (n ≠ -1): ∫ x^n dx = x^(n+1)/(n+1) + C.
  3. Special case (n = -1): ∫ (1/x) dx = ln|x| + C.
  4. Exponential and trig: ∫ e^x dx = e^x + C, ∫ sin x dx = -cos x + C, ∫ cos x dx = sin x + C.
  5. Linearity: ∫ (a f(x) + b g(x)) dx = a ∫ f(x) dx + b ∫ g(x) dx.
  6. Integration by substitution: If u = g(x), then ∫ f(g(x)) g'(x) dx = ∫ f(u) du.
📊 Visual ideas
Plot of y = f(x) (e.g., y = x^2) on [0,2] with the area under the curve shaded between x=0 and x=2. Label axes and show the slice rectangles (Riemann sum) converging to the shaded area.
Graph showing a function f(x) and several antiderivatives F(x)+C differing by vertical shifts to illustrate the constant of integration.
Plot of y = sin x with the region between the curve and the x-axis shaded from x = 0 to x = π to visualize ∫_0^π sin x dx = 2.
Illustration of substitution: plot y = cos(x^2) and indicate a small region and the change of variable u = x^2; show mapping of the region on the u-axis.
🔢2

Indefinite Integrals — Standard Formulas

What is an indefinite integral? An indefinite integral of a function f(x) is a function F(x) whose derivative is f(x). It is written as ∫f(x) dx = F(x) + C, where C is an arbitrary constant (the constant of integration). The indefinite integral represents the family of all antiderivatives of f.

Basic ideas and properties

  • Linearity: ∫[a f(x) + b g(x)] dx = a ∫ f(x) dx + b ∫ g(x) dx.
  • Power rule (inverse of differentiation): for n ≠ −1, ∫ x^n dx = x^{n+1}/(n+1) + C.
  • Special case: ∫ x^{-1} dx = ∫ (1/x) dx = ln|x| + C.
  • Substitution idea: If u = g(x) and du = g'(x) dx, then ∫ f(g(x)) g'(x) dx = ∫ f(u) du (useful for chain-rule reversal).
  • Interpretation: The indefinite integral gives a family of curves that differ by a vertical translation. In applied contexts it recovers accumulated quantity from a rate (e.g., position from velocity).

How to use the standard formulas: Recognize parts of the integrand that match a known formula, factor constants, split sums, and use substitution when you see a composition with a derivative factor. For products where substitution is not direct, use integration by parts: ∫ u dv = uv − ∫ v du.

📌 Examples
  • Example 1 — Power rule and logarithm: ∫ (3x^2 - 4/x + 5) dx Steps: integrate termwise. ∫ 3x^2 dx = 3 * (x^3/3) = x^3. ∫ ( -4/x) dx = -4 ∫ x^{-1} dx = -4 ln|x|. ∫ 5 dx = 5x. Answer: x^3 - 4 ln|x| + 5x + C.
  • Example 2 — Trigonometric substitution: ∫ sin(3x) dx Let u = 3x ⇒ du = 3 dx ⇒ dx = du/3. Then ∫ sin(3x) dx = (1/3) ∫ sin u du = -(1/3) cos u + C = -(1/3) cos(3x) + C.
  • Example 3 — Inverse trig formula: ∫ dx/(x^2 + 9) Recognize a^2 = 9, a = 3. ∫ dx/(x^2 + a^2) = (1/a) tan^{-1}(x/a) + C. So answer: (1/3) tan^{-1}(x/3) + C.
  • Example 4 — Substitution with exponential: ∫ 2 e^{5x} dx Let u = 5x ⇒ du = 5 dx ⇒ dx = du/5. Or use formula ∫ e^{ax} dx = (1/a) e^{ax}. So ∫ 2 e^{5x} dx = 2 * (1/5) e^{5x} + C = (2/5) e^{5x} + C.
  • Example 5 — Integration by parts: ∫ x e^x dx Choose u = x, dv = e^x dx ⇒ du = dx, v = e^x. Then ∫ x e^x dx = x e^x - ∫ e^x dx = x e^x - e^x + C = e^x(x - 1) + C.
🧮 Formulas
  1. Linearity: ∫[a f(x) + b g(x)] dx = a ∫ f(x) dx + b ∫ g(x) dx
  2. Constant: ∫ k dx = kx + C
  3. \[Power rule: ∫ x^n dx = x^{n+1}/(n+1) + C\]
    \[for n ≠ -1\]
  4. Log special case: ∫ dx/x = ln|x| + C
  5. \[Exponential: ∫ e^{ax} dx = (1/a) e^{ax} + C\]
  6. \[General exponential: ∫ a^{x} dx = a^{x}/ln a + C\]
    \[(a>0\]
    \[a≠1)\]
📊 Visual ideas
Graph idea 1: Plot y = x^2 and its antiderivative family y = x^3/3 + C on the same axes (x in [-2,2]). Show how changing C shifts the cubic vertically; area under parabola between x=a and x=b equals difference of antiderivative values.
Graph idea 2: Plot y = 1/x and y = ln|x| (x in [-5,-0.2] and [0.2,5]). Emphasize vertical asymptote at x=0 and that the antiderivative is defined on each sign-interval separately.
Graph idea 3: Plot y = sin x and y = -cos x (x in [-2π,2π]). Show phase relationship: derivative of -cos x is sin x. Display one member of family with different vertical shifts.
Graph idea 4: Plot y = sec^2 x and y = tan x (x in (-π/2,π/2) avoiding vertical asymptotes). Show that ∫ sec^2 x dx = tan x + C.
⚖️3

Methods of Integration

Integration is the inverse process of differentiation and is used to find areas, accumulated quantities and antiderivatives. 'Methods of Integration' refers to a set of techniques used to evaluate indefinite integrals (antiderivatives) and definite integrals when a direct antiderivative is not obvious.

Main methods (with when to use):

  • Substitution (u-substitution) — use when integrand is of the form f(g(x))·g'(x) so a change of variable simplifies the integral.
  • Integration by parts — useful for products of functions (polynomial·exponential, polynomial·log, polynomial·trig). Based on the product rule for derivatives.
  • Partial fraction decomposition — for rational functions (quotient of polynomials); decompose into simpler rational terms whose integrals are known.
  • Trigonometric integrals and identities — for integrands involving powers or products of sin x, cos x, tan x, etc.; use identities and substitutions like sin^2 x = (1−cos2x)/2.
  • Trigonometric substitution — for integrands with root expressions like sqrt(a^2 − x^2), sqrt(a^2 + x^2), sqrt(x^2 − a^2); substitute x = a sin θ, x = a tan θ, x = a sec θ respectively.
  • Weierstrass substitution (t = tan(x/2)) — converts rational functions of sin x and cos x into rational functions of t, useful for complicated trig integrals.
  • Reduction formulas — recursive relations for integrals of powers (e.g., ∫sin^n x dx) allowing reduction to lower powers.

Strategy: try to simplify the integrand (algebra/trig identities), look for a derivative inside the integrand (for substitution), choose parts wisely for integration by parts (let u be the function that simplifies when differentiated), and break rational functions into partial fractions when degree of numerator < denominator.

📌 Examples
  • 1) Substitution: ∫2x cos(x^2 + 1) dx Solution sketch: Let u = x^2 + 1 ⇒ du = 2x dx. Then integral becomes ∫cos u du = sin u + C = sin(x^2 + 1) + C.
  • 2) Integration by parts: ∫x e^x dx Solution sketch: Choose u = x (so du = dx), dv = e^x dx (so v = e^x). Then ∫x e^x dx = uv − ∫v du = x e^x − ∫e^x dx = x e^x − e^x + C = e^x(x − 1) + C.
  • 3) Partial fractions: ∫(2x + 3)/(x^2 + x − 2) dx Solution sketch: Factor denominator: x^2 + x − 2 = (x + 2)(x − 1). Decompose (2x+3)/[(x+2)(x−1)] = A/(x+2) + B/(x−1). Solve: 2x + 3 = A(x − 1) + B(x + 2). Comparing coefficients gives A = 1, B = 1. So integral = ∫(1/(x+2) + 1/(x−1)) dx = ln|x+2| + ln|x−1| + C.
  • 4) Trigonometric integral: ∫sin^3 x dx Solution sketch: Write sin^3 x = sin x(1 − cos^2 x). Let u = cos x ⇒ du = −sin x dx. Then ∫sin^3 x dx = ∫sin x(1 − cos^2 x) dx = −∫(1 − u^2) du = −(u − u^3/3) + C = −cos x + (cos^3 x)/3 + C.
  • 5) Real-life example (area/accumulation): If v(t) = 3t^2 (m/s) is the velocity of an object, the distance traveled from t = 0 to t = 2 s is ∫_0^2 3t^2 dt = [t^3]_0^2 = 8 m. Interpretation: integral accumulates instantaneous rate (velocity) to give total quantity (distance).
  • 6) Real-life example (work done by a spring): Force by a spring F(x) = kx (Hooke's law). Work to stretch from x=a to x=b is W = ∫_a^b kx dx = (k/2)(b^2 − a^2).
🧮 Formulas
  1. Substitution: If u = g(x) and du = g'(x) dx, then ∫f(g(x)) g'(x) dx = ∫f(u) du.
  2. Integration by parts: ∫u dv = u v − ∫v du (choose u to simplify on differentiation).
  3. Partial fractions (simple linear): P(x)/[(x−a)(x−b)] = A/(x−a) + B/(x−b); integrate termwise as ln|x−a| etc.
  4. Trigonometric identities: sin^2 x = (1 − cos 2x)/2, cos^2 x = (1 + cos 2x)/2, sin 2x = 2 sin x cos x.
  5. Trigonometric substitutions: For √(a^2 − x^2) use x = a sin θ; for √(a^2 + x^2) use x = a tan θ; for √(x^2 − a^2) use x = a sec θ.
  6. \[Standard integrals: ∫x^n dx = x^{n+1}/(n+1) + C (n ≠ −1), ∫e^x dx = e^x + C, ∫sin x dx = −cos x + C, ∫cos x dx = sin x + C, ∫dx/x = ln|x| + C, ∫dx/(x^2 + a^2) = (1/a) arctan(x/a) + C.\]
📊 Visual ideas
Plot integrand f(x) and its antiderivative F(x) on same axes (choose C so F(a)=0). Example: f(x)=2x cos(x^2+1) and F(x)=sin(x^2+1). Range: x in [−2,2]. This visualizes how slope of F equals f.
Area under curve: graph v(t)=3t^2 from t=0 to t=2 and shade area under curve to illustrate distance = ∫ v(t) dt.
Integration by parts visualization: plot x·e^x and components u=x and dv=e^x to show how splitting simplifies; also plot resulting antiderivative e^x(x−1). Range: x in [−1,2].
Partial fractions: plot rational function (2x+3)/(x^2+x−2) and separately plot the simpler components 1/(x+2) and 1/(x−1) to show how decomposition adds up. Use x avoiding vertical asymptotes (e.g., x∈[−5,5] excluding −2 and 1).
🔢4

Definite Integrals

Definition (Riemann sum): If f is a function defined on [a,b], its definite integral is the limit of Riemann sums (when the limit exists):

∫ab f(x) dx = limn→∞ Σi=1n f(xi*) Δx, where Δx = (b−a)/n and xi* ∈ [xi−1, xi].

Geometric meaning: The definite integral gives the signed area under the curve y = f(x) between x = a and x = b. Areas above the x-axis contribute positively, below contribute negatively.

Fundamental Theorem of Calculus (two parts):

  • Part I (Relationship of integral and derivative): If f is continuous on [a,b] and F(x) = ∫ax f(t) dt, then F is differentiable and F′(x) = f(x).
  • Part II (Evaluation): If F is any antiderivative of f on [a,b] (i.e. F′(x)=f(x)), then ∫ab f(x) dx = F(b) − F(a).

Basic properties (used often in computations):

  • Linearity: ∫ab [αf(x)+βg(x)] dx = α∫ab f(x) dx + β∫ab g(x) dx.
  • Additivity: ∫ab f(x) dx = ∫ac f(x) dx + ∫cb f(x) dx for a≤c≤b.
  • Reversal: ∫ba f(x) dx = −∫ab f(x) dx.
  • Zero width: ∫aa f(x) dx = 0.
  • Monotonicity: If f(x) ≥ g(x) on [a,b], then ∫ab f(x) dx ≥ ∫ab g(x) dx.
  • |∫ f| ≤ ∫ |f| (absolute value inequality).
  • Even/Odd functions (symmetric limits): If f is even, ∫−aa f(x) dx = 2∫0a f(x) dx. If f is odd, ∫−aa f(x) dx = 0.

Common evaluation techniques: Use an antiderivative (FTC II). Useful methods to find antiderivatives include substitution (change of variable) and integration by parts (definite form). For difficult integrals, numerical methods (Riemann sums, trapezoidal rule, Simpson's rule) approximate the value.

Mean Value Theorem for Integrals: If f is continuous on [a,b], there exists c in (a,b) such that ∫ab f(x) dx = f(c)(b−a). This gives the average value favg = (1/(b−a))∫ab f(x) dx.

Area between two curves: If f(x) ≥ g(x) on [a,b], area between them is ∫ab [f(x) − g(x)] dx.

Notes for CBSE Class 12: practice evaluating definite integrals by finding antiderivatives, using properties to simplify limits or symmetry, and interpreting integrals in application problems (area, displacement, probability).

📌 Examples
  • Displacement from velocity: If v(t) is velocity, displacement from t1 to t2 is ∫_{t1}^{t2} v(t) dt. If speed is needed, integrate |v(t)|.
  • Area between curves: Area between y = f(x) and y = g(x) from x=a to b (with f≥g) is ∫_{a}^{b} [f(x) − g(x)] dx. Useful for finding land area, engineering cross-sections.
  • Work by a variable force: If a force F(x) acts along a line as an object moves from x=a to x=b, work = ∫_{a}^{b} F(x) dx.
  • Probability with a continuous PDF: If f(x) is a probability density on [a,b], probability that X lies in [c,d] ⊆ [a,b] is ∫_{c}^{d} f(x) dx.
  • Volume by slicing: Volume of a solid with cross-sectional area A(x) perpendicular to x over [a,b] is V = ∫_{a}^{b} A(x) dx (used in tanks, engineering shapes).
🧮 Formulas
  1. \[Definition (Riemann): ∫_{a}^{b} f(x) dx = lim_{n→∞} Σ_{i=1}^{n} f(x_i^*) Δx, Δx=(b−a)/n.\]
  2. \[Fundamental Theorem (evaluation): If F'(x)=f(x) then ∫_{a}^{b} f(x) dx = F(b) − F(a).\]
  3. \[Substitution (change of variable): If x = φ(t)\]
    \[dx = φ'(t) dt and x limits map to t limits\]
    \[then ∫_{a}^{b} f(x) dx = ∫_{α}^{β} f(φ(t)) φ'(t) dt.\]
  4. \[Integration by parts (definite): ∫_{a}^{b} u dv = [u v]_{a}^{b} − ∫_{a}^{b} v du.\]
  5. \[Additivity: ∫_{a}^{b} f = ∫_{a}^{c} f + ∫_{c}^{b} f.\]
  6. \[Reversal: ∫_{b}^{a} f = −∫_{a}^{b} f.\]
📊 Visual ideas
Graph y = f(x) on [a,b] with the region between the curve and x-axis from x=a to x=b shaded (showing positive area when f≥0 and negative when f≤0).
Riemann-sum illustration: partition [a,b] into n subintervals, draw sample rectangles (left/right/midpoint) to show approximation to area and convergence as n increases.
Antiderivative vs function: plot f(x) and its antiderivative F(x) on the same axes; show that slope of F equals f (illustrates FTC Part I).
Even/odd symmetry: plot an even function and shade symmetric areas on [−a,a] to show doubling; plot an odd function to show cancellation (areas cancel to zero).
🔢5

Properties and Symmetry in Definite Integrals

Definite integrals have algebraic properties that make evaluation easier, especially when the integrand or interval has symmetry. Key ideas: linearity, additivity of intervals, change of limits sign, and symmetry about the origin or about the mid-point of the interval. Using symmetry often reduces an integral over a large interval to an integral over half the interval or sometimes shows the integral is zero.

Important symmetry types:

  • Even functions: f(−x) = f(x). For any a > 0, ∫−aa f(x) dx = 2 ∫0a f(x) dx. Geometrically the area on the left and right of the y-axis are equal.
  • Odd functions: f(−x) = −f(x). For any a, ∫−aa f(x) dx = 0. The positive area cancels the negative area because of antisymmetry.
  • Mid-point or interval symmetry: On [a,b], many integrals simplify using the substitution x → a + b − x. In particular ∫ab f(x) dx = ∫ab f(a+b−x) dx, so sometimes f(x)+f(a+b−x) is simpler (or constant).

Why this matters: symmetry reduces computation and gives insight (for example, center-of-mass or expected value of a symmetric distribution often becomes zero). Combining properties with substitution or algebraic manipulation can turn a difficult integral into an easy one.

Practical tips:

  • Check if f is even or odd before integrating on [−a,a].
  • For integrals on [a,b] try the substitution x → a+b−x to detect mid-point symmetry.
  • If f(x)+f(a+b−x) simplifies to a constant or simple function, evaluate the average times interval length.
📌 Examples
  • Example 1 — Odd function: Evaluate I = ∫_{−2}^{2} x^3 dx. Since x^3 is odd, I = 0.
  • Example 2 — Even function: Evaluate I = ∫_{−π/2}^{π/2} cos x dx. cos x is even, so I = 2 ∫_{0}^{π/2} cos x dx = 2[sin x]_{0}^{π/2} = 2.
  • Example 3 — Mid-point symmetry: Evaluate I = ∫_{0}^{π} sin x dx. Note sin(π − x) = sin x, so I = 2 ∫_{0}^{π/2} sin x dx = 2[−cos x]_{0}^{π/2} = 2(1) = 2.
  • Example 4 — Product of even and odd: Evaluate I = ∫_{−3}^{3} x^2 sin x dx. x^2 is even, sin x is odd, product is odd, so I = 0.
🧮 Formulas
  1. \[Linearity: ∫_{a}^{b} [α f(x) + β g(x)] dx = α ∫_{a}^{b} f(x) dx + β ∫_{a}^{b} g(x) dx\]
  2. \[Additivity of interval: ∫_{a}^{b} f(x) dx = ∫_{a}^{c} f(x) dx + ∫_{c}^{b} f(x) dx for any c between a and b\]
  3. \[Reversal of limits: ∫_{a}^{b} f(x) dx = − ∫_{b}^{a} f(x) dx\]
  4. \[Even function (f(−x)=f(x)): ∫_{−a}^{a} f(x) dx = 2 ∫_{0}^{a} f(x) dx\]
  5. \[Odd function (f(−x)=−f(x)): ∫_{−a}^{a} f(x) dx = 0\]
  6. \[Midpoint symmetry: ∫_{a}^{b} f(x) dx = ∫_{a}^{b} f(a+b−x) dx\]
📊 Visual ideas
Plot y = x^2 on [−2,2]: show symmetric areas about the y-axis; shade the area on [0,2] and mirror it to demonstrate ∫_{−2}^{2} x^2 dx = 2 ∫_{0}^{2} x^2 dx.
Plot y = x^3 on [−2,2]: show positive area for x>0 and equal-magnitude negative area for x<0 to visualize why the integral is zero.
Plot y = sin x on [0,π]: mark x = π/2 vertical line and shade left and right halves to illustrate midpoint symmetry sin(π − x) = sin x and using 2 ∫_{0}^{π/2}.
Plot a generic function f(x) on [a,b] and plot its reflection f(a+b−x) on the same axes; highlight that adding the two values at same x gives symmetrical simplification—use example f(x)=x and interval [0,2] so f(x)+f(2−x)=2 (constant), then shade to show integral equals constant*(b−a)/2.
🔢6

Techniques for Evaluating Definite Integrals

Overview. A definite integral ∫_a^b f(x) dx gives the signed area under the curve y = f(x) from x = a to x = b. Key techniques transform a given integral into an easier one or exploit properties of the integrand or limits.

Main techniques.

  • Fundamental Theorem of Calculus (FTC). If F is an antiderivative of f, then ∫_a^b f(x) dx = F(b) − F(a). Use FTC after finding an antiderivative.
  • Substitution (Change of Variable). If x = g(t) is a smooth bijection on the interval, transform ∫_a^b f(x) dx to ∫_{g^{-1}(a)}^{g^{-1}(b)} f(g(t)) g'(t) dt. This simplifies integrands that are compositions.
  • Integration by Parts for Definite Integrals. Use ∫_a^b u dv = [u v]_a^b − ∫_a^b v du to reduce products of functions, e.g., x e^x, x sin x.
  • Symmetry and the Property f(x) ↔ f(a+b−x). ∫_a^b f(x) dx = ∫_a^b f(a+b−x) dx. For a = −b and even/odd functions: if f is even, ∫_{−a}^{a} f(x) dx = 2 ∫_0^a f(x) dx; if f is odd, ∫_{−a}^{a} f(x) dx = 0.
  • Decomposition and Partial Fractions. Break rational functions into simpler fractions whose antiderivatives are known.
  • Reduction Formulas. Use recurrence relations to reduce powers or parameters step by step.
  • Use of Definite Integral Properties. Linearity, additivity of intervals, reversal of limits, and comparison inequalities help simplify or estimate integrals.
  • Known Standard Integrals. Recognize forms like ∫ 1/(1+x^2) dx = arctan x, ∫ e^x dx = e^x, ∫ sin x dx = −cos x, etc.
  • Numerical Methods (if not integrable in closed form). Trapezoidal rule and Simpson's rule approximate definite integrals when antiderivatives are hard or impossible to find.

Strategy to choose a technique. Look for substitution patterns (inner derivative present), product forms (use parts), symmetry of interval or integrand (use even/odd or f(a+b−x)), rational functions (try partial fractions), and standard integral shapes.

Important remarks. For improper integrals (infinite limits or singularities), treat as limits: ∫_a^{∞} f(x) dx = lim_{R→∞} ∫_a^R f(x) dx, and check convergence.

📌 Examples
  • 1) Substitution: Evaluate ∫_0^1 (2x)/(1+x^2)^2 dx. Let u = 1 + x^2 ⇒ du = 2x dx. Limits: u(0)=1, u(1)=2. Integral = ∫_1^2 u^{−2} du = [−1/u]_1^2 = −1/2 + 1 = 1/2.
  • 2) Integration by parts: Evaluate ∫_0^π x sin x dx. Let u = x ⇒ du = dx; dv = sin x dx ⇒ v = −cos x. Then ∫ u dv = [u v]_0^π − ∫_0^π v du = [−x cos x]_0^π + ∫_0^π cos x dx = (−π cos π + 0) + (sin x)_0^π = −π(−1) = π.
  • 3) Symmetry/substitution: Evaluate ∫_0^π sin x /(1+cos^2 x) dx. Let u = cos x ⇒ du = −sin x dx. Limits: u(0)=1, u(π)=−1. Integral = ∫_1^{−1} −1/(1+u^2) du = ∫_{−1}^1 1/(1+u^2) du = [arctan u]_{−1}^1 = arctan 1 − arctan(−1) = π/4 + π/4 = π/2.
  • 4) Even/odd function: Evaluate ∫_{−2}^2 x^3 cos x dx. Note x^3 cos x is odd (product of odd and even = odd). For symmetric limits, integral = 0.
  • 5) Decomposition: Evaluate ∫_0^1 (1)/(x^2+x) dx. Factor denominator = x(x+1). Partial fractions: 1/(x(x+1)) = 1/x − 1/(x+1). Integrate from 0 to 1 (improper at 0 treat as limit): ∫_ε^1 (1/x − 1/(x+1)) dx = [ln x − ln(x+1)]_ε^1 = (ln1 − ln2) − (ln ε − ln(ε+1)). Taking ε→0+, ln ε → −∞ but ln(ε+1)→0, the ln ε terms cancel when combined properly with correct limit, giving finite value: ln(1)−ln2 − (−∞−0) — actually handle by recognizing integrand is integrable on (0,1], result = ln2. (Alternate direct antiderivative: ln(x/(x+1)), evaluate carefully: ln(1/2) − lim_{x→0+} ln(0)= finite after limit = ln2.)
  • 6) Using f(a+b−x): If I = ∫_0^1 ln(1+x)/(1+x^2) dx, sometimes pairing I with itself using x→1−x can simplify; such symmetry tricks often reduce integrals to simpler forms or standard constants.
🧮 Formulas
  1. Linearity: ∫_a^b [α f(x)+β g(x)] dx = α ∫_a^b f(x) dx + β ∫_a^b g(x) dx
  2. Additivity: ∫_a^b f(x) dx = ∫_a^c f(x) dx + ∫_c^b f(x) dx
  3. Reversal: ∫_b^a f(x) dx = −∫_a^b f(x) dx
  4. Fundamental Theorem: If F'(x)=f(x), then ∫_a^b f(x) dx = F(b) − F(a)
  5. \[Substitution: If x = g(t)\]
    \[dx = g'(t) dt\]
    \[then ∫_a^b f(x) dx = ∫_{g^{-1}(a)}^{g^{-1}(b)} f(g(t)) g'(t) dt\]
  6. Integration by parts (definite): ∫_a^b u dv = [u v]_a^b − ∫_a^b v du
📊 Visual ideas
Plot y = x^2 on [0,2] and shade the area under curve from x=0 to x=2 to illustrate ∫_0^2 x^2 dx. Show F(x) = x^3/3 as antiderivative evaluated at endpoints.
Plot an odd function like y = x^3 and shade symmetric area from −2 to 2 to show cancellation leading to integral 0.
Plot y = sin x/(1+cos^2 x) on [0,π] and indicate substitution u = cos x mapping x-axis values to u from 1 to −1; shade area to illustrate change of variable.
Sketch product functions like y = x e^x on [0,1] and illustrate the use of integration by parts by relating area pieces to boundary terms [x e^x]_0^1 and the integral of e^x.
🟦7

Applications of Integrals — Area

Introduction
The definite integral gives the net signed area under a curve. Using limits of Riemann sums, area of a plane region bounded by curves can be computed by evaluating definite integrals. In Class 12 you use integrals to find:

  • Area under a single curve and the x-axis between x = a and x = b.
  • Area between two curves y = f(x) and y = g(x) over [a, b].
  • Area computed by integrating with respect to y when convenient.

Derivation idea (brief)
Split the interval [a,b] into n small subintervals of width Δx. Approximate the area by summing areas of rectangles of height f(x_i*) and base Δx. Taking the limit as n → ∞ gives the definite integral A = ∫_a^b f(x) dx (if f(x) ≥ 0 on [a,b]).

Sign consideration
A definite integral ∫_a^b f(x) dx gives signed area: parts above the x-axis contribute positive area, parts below contribute negative. For geometric area always take absolute values or split intervals where sign changes.

Area between two curves
If f(x) ≥ g(x) on [a,b], the area of the region between them is
A = ∫_a^b [f(x) − g(x)] dx. If f and g cross within [a,b], find intersection points, split the integral at those x-values, and add absolute values of each piece.

Integrating with respect to y
When curves are better described as x = φ(y), x = ψ(y) on y ∈ [c,d], the area between them is
A = ∫_c^d [x_right(y) − x_left(y)] dy. This is useful when vertical strips would require splitting but horizontal strips give a single integral.

Procedure / Steps to find area between curves

  • Sketch the curves and find intersection points (solve f(x)=g(x)).
  • Decide which function is on top (greater) on each subinterval.
  • Set up the integral(s): A = ∑ ∫ [top − bottom] dx (or dy).
  • Evaluate the integral(s) and add results (use absolute values if needed).

Common mistakes to avoid

  • Not checking which function is greater on the interval.
  • Using signed integral value directly when asked for geometric area (use absolute values or split intervals where sign changes).
  • For regions bounded by vertical and non-vertical lines, choose dx or dy appropriately to avoid unnecessary splitting.

📌 Examples
  • Example 1 — Area under a curve: Find area under y = x^2 from x = 0 to x = 2. Solution: A = ∫_0^2 x^2 dx = [x^3/3]_0^2 = 8/3 square units.
  • Example 2 — Area between two curves: Find area between y = x and y = x^2 from x = 0 to x = 1. Solution: On [0,1], x ≥ x^2, so A = ∫_0^1 (x − x^2) dx = [x^2/2 − x^3/3]_0^1 = 1/2 − 1/3 = 1/6 square units.
  • Example 3 — Using dy: Find area between x = y and x = y^2 from y = 0 to y = 1. Solution: For y in [0,1], x_right = y, x_left = y^2. A = ∫_0^1 (y − y^2) dy = [y^2/2 − y^3/3]_0^1 = 1/2 − 1/3 = 1/6 square units.
🧮 Formulas
  1. Area under curve above x-axis: A = ∫_a^b f(x) dx, if f(x) ≥ 0 on [a,b].
  2. If f(x) ≤ 0 on [a,b], geometric area = −∫_a^b f(x) dx.
  3. Area between curves y = f(x) and y = g(x) (f ≥ g): A = ∫_a^b [f(x) − g(x)] dx.
  4. General (no order assumed): A = ∫_a^b |f(x) − g(x)| dx or split at intersections.
  5. Area using horizontal strips: A = ∫_c^d [x_right(y) − x_left(y)] dy.
  6. \[When curves cross: find intersection points x_1 < x_2 < ... and compute A = Σ ∫_{x_k}^{x_{k+1}} |f(x) − g(x)| dx.\]
📊 Visual ideas
Plot 1: y = x^2 on x ∈ [0,2] with the area under the curve shaded between x=0 and x=2. Show vertical sample rectangles (Riemann strips) to illustrate the integral approximation. Label area = ∫_0^2 x^2 dx.
Plot 2: y = x and y = x^2 on x ∈ [−0.5,1.5]. Shade the region between the curves from x=0 to x=1. Mark intersection points (0,0) and (1,1), and indicate 'top' and 'bottom' functions.
Plot 3: y = sin x over x ∈ [0,π]. Shade the area between the curve and the x-axis. Label the exact area 2 (since ∫_0^π sin x dx = 2).
Plot 4 (use dy): x = y and x = y^2 for y ∈ [0,1]. Draw horizontal strips and shade the region between the curves. Mark x_right=y and x_left=y^2 and write the integral A = ∫_0^1 (y − y^2) dy.
🔢8

Special Forms and Examples

What are Special Forms? In integration, certain integrands appear in standard patterns whose antiderivatives are immediate once you recognise the pattern. These are called special forms. They typically arise when the integrand contains a function and its derivative (or a constant multiple of it), or matches known elementary integral formulas (logarithmic, inverse trigonometric, exponential, trigonometric).

How to use them: Look for an inner function u = f(x) whose derivative f'(x) (or a constant multiple) is present in the integrand. Use substitution u = f(x) to convert the integral into a basic form. Always add the constant of integration C for indefinite integrals.

Common patterns (idea):

  • If integrand is f'(x)/f(x) → result ln|f(x)| + C.
  • If integrand is f'(x) (f(x))^n → result (f(x))^{n+1}/(n+1) + C for n ≠ -1.
  • If integrand is f'(x) e^{f(x)} → result e^{f(x)} + C.
  • If integrand is f'(x) [sin f(x)] or [cos f(x)] → use substitution to get cos or -sin forms.
  • Standard trigonometric and inverse-trigonometric forms and rational forms (partial fractions) should be memorised.

Remarks:

  • Constant multiple: If you have k f'(x)/f(x), factor out k: k ln|f(x)| + C.
  • Not every integrand has an elementary antiderivative (e.g. e^{-x^2} has no elementary primitive; its definite integral is handled via Gaussian methods or special functions).
📌 Examples
  • Example 1: ∫ 2x/(x^2 + 1) dx. Let u = x^2 + 1 ⇒ du = 2x dx. So ∫ du/u = ln|u| + C = ln(x^2 + 1) + C.
  • Example 2: ∫ 3x^2 (x^3 + 1)^4 dx. Let u = x^3 + 1 ⇒ du = 3x^2 dx. So integral = ∫ u^4 du = u^5/5 + C = (x^3 + 1)^5/5 + C.
  • Example 3: ∫ e^{3x} dx. Let u = 3x ⇒ du = 3 dx ⇒ dx = du/3. So integral = (1/3) ∫ e^u du = (1/3) e^{3x} + C.
  • Example 4: ∫ dx/(x^2 + a^2) = (1/a) arctan(x/a) + C. (Standard inverse-trig form.)
  • Example 5: ∫ x / √(x^2 + 4) dx. Let u = x^2 + 4 ⇒ du = 2x dx ⇒ x dx = du/2. Then integral = ∫ (1/2) u^{-1/2} du = (1/2) * 2 u^{1/2} + C = √(x^2 + 4) + C.
  • Example 6 (trig): ∫ sin x cos x dx. Use u = sin x ⇒ du = cos x dx ⇒ ∫ u du = u^2/2 + C = (sin^2 x)/2 + C.
🧮 Formulas
  1. ∫ f'(x)/f(x) dx = ln|f(x)| + C
  2. \[∫ f'(x) [f(x)]^n dx = [f(x)]^{n+1}/(n+1) + C (n ≠ -1)\]
  3. \[∫ f'(x) e^{f(x)} dx = e^{f(x)} + C\]
  4. ∫ f'(x) sin(f(x)) dx = -cos(f(x)) + C (up to constant sign by substitution)
  5. ∫ dx/(ax + b) = (1/a) ln|ax + b| + C
  6. ∫ dx/(x^2 + a^2) = (1/a) arctan(x/a) + C
📊 Visual ideas
Plot y = 1/x for x ≠ 0 and show that area under 1/x from 1 to t corresponds to ln t (label the area ≈ ∫_1^t 1/x dx = ln t).
Plot y = 1/(x^2 + a^2) (choose a = 1) and show the bell-like curve; annotate that its antiderivative is arctan(x) and mark arctan values at several x.
Plot y = e^{3x} and show how vertical scaling changes the integral: area under curve from x0 to x1 equals (1/3)(e^{3x1} - e^{3x0}).
Plot y = sin x · cos x over one period; show substitution u = sin x turns area into (1/2) sin^2 x. Mark corresponding primitive curve (sin^2 x / 2).

Key Concepts

Integral
A mathematical object representing accumulation; can be indefinite (antiderivative) or definite (signed area/limit of sums).
Antiderivative
A function F(x) whose derivative is f(x); i.e., F'(x) = f(x).
Indefinite integral
The family of all antiderivatives of f(x), written ∫ f(x) dx = F(x) + C, where C is an arbitrary constant.
Definite integral
The limit of Riemann sums giving the net signed area from a to b: ∫_a^b f(x) dx (a and b are limits).
Riemann sum
A finite sum Σ f(x_i*) Δx approximating area; the definite integral is its limit as partition mesh → 0.
Fundamental Theorem of Calculus
Links differentiation and integration: (1) If F(x)=∫_a^x f(t) dt then F'(x)=f(x). (2) If F is an antiderivative of f, ∫_a^b f(x) dx = F(b)-F(a).
Integration by substitution
Change of variable method: set u = g(x) to simplify the integrand and integrate in u.
Integration by parts
Formula ∫ u dv = uv - ∫ v du used to integrate products of functions.
Trigonometric integrals
Integrals involving trigonometric functions solved using identities, substitution, or reduction formulas.
Rational function
A quotient of two polynomials. Integrals often require algebraic manipulation (division/partial fractions).
Partial fraction decomposition
Expressing a rational function as a sum of simpler fractions to integrate termwise.
Reduction formula
A recurrence relation expressing an integral with parameter n in terms of an integral with lower parameter (n−1 or n−2).
Properties of definite integrals
Linearity and additivity: ∫_a^b [αf(x)+βg(x)] dx = α∫_a^b f + β∫_a^b g; ∫_a^b f = ∫_a^c f + ∫_c^b f.
Even and odd function symmetry
If f is even, ∫_{-a}^a f(x) dx = 2∫_0^a f(x) dx. If f is odd, ∫_{-a}^a f(x) dx = 0 (when integrable).
Area under a curve
If f(x) ≥ 0 on [a,b], the area between curve y=f(x), x-axis and x=a,b is ∫_a^b f(x) dx.
Change of limits (definite integrals)
When substituting u=g(x) in ∫_a^b f(x) dx, update limits: x=a→u=g(a), x=b→u=g(b).
Improper integral
An integral with infinite interval or unbounded integrand, evaluated as a limit of proper integrals.
Average value of a function
Average on [a,b] is (1/(b-a)) ∫_a^b f(x) dx.
Standard integral forms
Basic antiderivatives frequently used: power, exponential, trigonometric and reciprocal forms.
Inverse trigonometric integrals
Integrals that yield inverse trig functions, often from rational functions with quadratic denominators.

End-of-Chapter Trial Paper & Test Questions

Topic-wise questions to test your understanding of every concept in this chapter.

  1. Define the indefinite integral and state why an arbitrary constant C appears. / अनिश्चित समाकल को परिभाषित कीजिए और बताइए कि स्वेच्छ अचर C क्यों आता है।
    Show answer

    ∫f(x)dx = F(x)+C where F'(x)=f(x); since any two antiderivatives of f differ only by a constant, C represents the whole family of antiderivatives. / ∫f(x)dx = F(x)+C जहाँ F'(x)=f(x); f के कोई भी दो प्रतिअवकलज केवल एक अचर से भिन्न होते हैं, इसलिए C प्रतिअवकलजों के समस्त कुल को दर्शाता है।

  2. Evaluate ∫(3x² − 4/x + 5)dx. / ∫(3x² − 4/x + 5)dx का मान ज्ञात कीजिए।
    Show answer

    Integrate termwise: x³ − 4 ln|x| + 5x + C. / पदशः समाकलन: x³ − 4 ln|x| + 5x + C।

  3. Using substitution, evaluate ∫2x cos(x²)dx. / प्रतिस्थापन से ∫2x cos(x²)dx का मान ज्ञात कीजिए।
    Show answer

    Let u=x², du=2x dx, so ∫cos u du = sin u + C = sin(x²)+C. / माना u=x², du=2x dx, अतः ∫cos u du = sin u + C = sin(x²)+C।

  4. State the Fundamental Theorem of Calculus (evaluation part) and use it to find ∫₀² x² dx. / कलन के आधारभूत प्रमेय (मूल्यांकन भाग) को लिखिए तथा ∫₀² x² dx ज्ञात कीजिए।
    Show answer

    If F'=f then ∫ₐᵇ f dx = F(b)−F(a); ∫₀² x² dx = [x³/3]₀² = 8/3. / यदि F'=f तो ∫ₐᵇ f dx = F(b)−F(a); ∫₀² x² dx = [x³/3]₀² = 8/3।

  5. Evaluate ∫x eˣ dx using integration by parts. / खण्डशः समाकलन द्वारा ∫x eˣ dx का मान ज्ञात कीजिए।
    Show answer

    Take u=x, dv=eˣdx; ∫x eˣ dx = x eˣ − ∫eˣ dx = eˣ(x−1)+C. / माना u=x, dv=eˣdx; ∫x eˣ dx = x eˣ − ∫eˣ dx = eˣ(x−1)+C।

  6. Evaluate ∫(2x+3)/(x²+x−2)dx by partial fractions. / आंशिक भिन्नों द्वारा ∫(2x+3)/(x²+x−2)dx ज्ञात कीजिए।
    Show answer

    x²+x−2=(x+2)(x−1); decomposition gives A=1, B=1, so integral = ln|x+2| + ln|x−1| + C. / x²+x−2=(x+2)(x−1); वियोजन से A=1, B=1, अतः समाकल = ln|x+2| + ln|x−1| + C।

  7. Using symmetry, evaluate ∫₋₂² x³ dx and justify the result. / सममिति का उपयोग कर ∫₋₂² x³ dx ज्ञात कीजिए और परिणाम का औचित्य दीजिए।
    Show answer

    x³ is odd, and for an odd function ∫₋ₐᵃ f dx = 0, so the integral = 0. / x³ विषम है, और विषम फलन के लिए ∫₋ₐᵃ f dx = 0, अतः समाकल = 0।

  8. Evaluate ∫₀^π x sin x dx. / ∫₀^π x sin x dx का मान ज्ञात कीजिए।
    Show answer

    By parts with u=x, dv=sin x dx: [−x cos x]₀^π + ∫₀^π cos x dx = π + [sin x]₀^π = π. / खण्डशः u=x, dv=sin x dx से: [−x cos x]₀^π + ∫₀^π cos x dx = π + [sin x]₀^π = π।

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