Definite integrals: fundamental theorem, symmetry and area

Goals and prerequisites

Evaluate definite integrals, distinguish signed accumulation from geometric area, and differentiate accumulation functions. Prerequisites: antiderivatives and continuity. Ordinary definite integrals below use real Riemann integration unless explicitly labelled improper.

1. From sums to an integral

Partition [a,b] into small pieces of width Δxᵢ, choose a sample ξᵢ in each piece and form Σf(ξᵢ)Δxᵢ. If these sums approach the same value as the maximum piece width tends to zero, independently of samples and partitions, that value is ∫ₐᵇ f(x)dx. A continuous function on a closed bounded interval is integrable. Bounded piecewise-continuous functions with finitely many discontinuities are also integrable.

The integral is signed accumulation: contributions below the axis are negative. The geometric area between the graph and axis is ∫ₐᵇ |f(x)|dx, obtained by splitting at sign changes. For f(x)=x on [−1,1], the signed integral is 0 but the geometric area is 1.

Linearity holds, ∫ₐᵃ f=0, reversing limits changes the sign, and intervals add: ∫[a,b] f(x)dx + ∫[b,c] f(x)dx = ∫[a,c] f(x)dx where the integrals exist. If m≤f≤M on [a,b], then m(b−a)≤∫ₐᵇ f≤M(b−a).

2. Fundamental theorem of calculus (FTC)

For continuous f on [a,b], define F(x)=∫ₐˣ f(t)dt. At interior x, F′(x)=f(x). An increment in accumulated area, divided by its width, approaches the local height. The letter t is a dummy variable, not the upper-limit variable x.

If G′=f and f is continuous on [a,b], then ∫ₐᵇ f(x)dx=G(b)−G(a). Thus ∫₀² (3x²+1)dx=[x³+x]₀²=10. Do not append +C to the numerical definite-integral result; an antiderivative constant cancels in the subtraction.

With differentiable variable limits and continuous integrand, the derivative of the integral from u(x) to v(x) of f(t)dt is f(v(x))v′(x)−f(u(x))u′(x). For H(x)=∫₁ˣ² ln t dt, x>0, the derivative is ln(x²)·2x. Restricting x>0 avoids a path to the singular point t=0. For K(x)=∫ₓ²ˣ sin t dt, K′=2sin(2x)−sin x.

3. Definite substitution and symmetry

For continuous f and continuously differentiable g on [a,b], ∫ₐᵇ f(g(x))g′(x)dx equals the integral of f(u)du from g(a) to g(b), with f continuous on an interval containing g([a,b]). When substituting, change both limits and the differential, or return fully to x before using the original limits.

Example: ∫₀¹ 2x/(1+x²)dx, with u=1+x², has new limits 1 and 2, so it is ∫₁² du/u=ln 2. For continuously differentiable u and v on [a,b], definite integration by parts gives: ∫ₐᵇ u v′=[uv]ₐᵇ−∫ₐᵇ v u′. Thus ∫₀¹ xeˣdx=[(x−1)eˣ]₀¹=1.

For integrable even f, ∫₋ₐᵃ f=2∫₀ᵃ f; for integrable odd f it is 0 (a>0). The integrability condition matters. Reflection also gives ∫₀ᵃ f(x)dx=∫₀ᵃ f(a−x)dx. To evaluate I=∫₀¹ x/(1+x²)dx, substitution gives (1/2)ln 2; an odd-looking numerator alone does not make an integral over [0,1] vanish.

4. Piecewise signs and area

To find area under |x−1| on [0,2], split at 1: ∫₀¹(1−x)dx+∫₁²(x−1)dx=1/2+1/2=1. The signed integral of x−1 over the same interval is zero. If two curves cross, area between them is the integral of the absolute difference, split at crossings; do not use one fixed upper-minus-lower formula across a sign change.

5. Improper integrals: a domain safeguard

If the integrand is unbounded at an endpoint, define the integral by a one-sided limit rather than blindly applying FTC across the singularity. For example, ∫₀¹ x^(−1/2)dx=lim(ε→0+) [2√x]ε¹=2, whereas ∫₀¹ dx/x=lim(ε→0+) (−ln ε)=+∞ and does not converge.

At an interior singularity, both sides must converge separately. ∫₋₁¹ dx/x does not converge as an improper integral: the left integral diverges to −∞ and the right to +∞. Symmetric cancellation gives a Cauchy principal value of zero, a different concept, not a convergent ordinary integral. This safeguard prevents misuse of odd symmetry and antiderivative subtraction.

Checklist: inspect singularities, identify signed integral versus area, use valid antiderivatives, transform limits consistently, include chain factors for variable bounds, and apply symmetry only after checking integrability.

Original practice: six single-correct MCQs

These are original questions, not previous-year questions. Choose exactly one option.

CF01. Evaluate ∫₀²(3x²+1)dx.

A. 8 B. 9 C. 10 D. 12

Answer: C

A: This omits the integral of 1.

B: The constant term contributes 2, so the total is 8+2=10, not 9.

C: Use [x³+x]₀²=8+2.

D: Direct antiderivative evaluation gives 10.

CF02. What is the geometric area between y=x and the x-axis over [−1,1]?

A. 0 B. 1/2 C. 2 D. 1

Answer: D

A: Zero is the signed integral.

B: This counts only one of two equal triangles.

C: Each triangle has area 1/2, not 1.

D: Integrating |x| gives 1/2+1/2=1.

CF03. For x>0, find d/dx ∫₁ˣ² ln t dt.

A. 2x ln(x²) B. ln(x²) C. 2/x D. ln x

Answer: A

A: FTC gives ln(x²), then the upper-limit derivative supplies 2x.

B: This omits the chain factor 2x.

C: This differentiates ln(x²), rather than the accumulation function.

D: The integrand must be evaluated at x² with its chain factor.

CF04. Evaluate ∫₀¹ 2x/(1+x²)dx.

A. 0 B. ln 2 C. 2ln 2 D. 1/2

Answer: B

A: The integrand is positive except at zero.

B: u=1+x² changes the bounds to 1 and 2.

C: The differential already absorbs 2x.

D: The antiderivative is ln(1+x²), not x²/2.

CF05. As an ordinary improper integral, ∫₋₁¹ dx/x:

A. Equals 0 / 0 है B. Equals 1 / 1 है C. Does not converge / अभिसरित नहीं D. Equals ln 2 / ln 2 है

Answer: C

A: Zero is only the symmetric principal value; the side integrals diverge.

B: The integral fails the separate-side convergence test.

C: At zero the left and right integrals diverge separately.

D: Antiderivative subtraction across the singularity is invalid.

CF06. Evaluate the improper integral ∫₀¹ x^(−1/2)dx.

A. 1/2 B. 1 C. Diverges / अपसारी D. 2

Answer: D

A: The antiderivative coefficient is 2, not 1/2.

B: Evaluate 2√x, not √x.

C: An unbounded integrand can have a convergent improper integral.

D: lim(ε→0+)2(1−√ε)=2.

Source and scope

Second Integration lesson: definite integration and essential singularity safeguards. Improper examples prevent invalid use of FTC; no claim that every integration technique is covered.

Official GATE 2026 CS syllabus (verified 2026-10-14).