2. Calculus — Integration

Evaluation of definite integrals

Definite integrals represent the net accumulated area under a curve between two fixed limits, rigorously defined via Riemann sums and evaluated using the Fundamental Theorem of Calculus. This topic covers definite integral evaluation, along with its powerful symmetry and reduction properties,.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade

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Definition and statement

Fundamental Theorem of Calculus (Parts I and II) & Definite Integral Definition

Let $f: [a, b] \to \mathbb{R}$ be continuous, and let $F(x) = \int_a^x f(t)\,dt$ for $x \in [a, b]$. Then $F$ is differentiable on $(a, b)$ and $F'(x) = f(x)$. Furthermore, if $G$ is any antiderivative of $f$ on $[a, b]$, then $$\int_a^b f(x)\,dx = G(b) - G(a)$$.

📐 Geometric Interpretation

Geometrically, $\int_a^b f(x)\,dx$ computes the net signed area bounded by the curve $y = f(x)$, the $x$-axis, and the vertical lines $x = a$ and $x = b$, where areas above the $x$-axis are positive and those below are negative.

⚙️ Algebraic & Structural Meaning

It bridges infinitesimal accumulation (sums of infinite terms) with anti-differentiation, transforming complex limit-of-sum problems into straightforward algebraic evaluations.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Continuity of Integrand	`[a, b] (Closed interval)`	Strictly Mandatory	If $f$ has essential or jump discontinuities, the standard Riemann integral may not exist, or the antiderivative relationship fails unless handled via Lebesgue integration or improper integrals.
Existence of Antiderivative	`(a, b) (Open interior) and endpoints`	Strictly Mandatory	Without an explicit elementary or special-function antiderivative, evaluation requires numerical quadrature, series expansion, or contour integration techniques.
Finiteness of Limits	`Finite bounds $a$ and $b$`	Essential for standard definite integrals	Infinite intervals ($a = -\infty$ or $b = \infty$) or unbounded integrands require conversion to improper integrals via limits.

Useful results and shortcuts

Use these identities and results when their conditions apply:

King's Property (Symmetric Integration)

$$\int_a^b f(x)\,dx = \int_a^b f(a + b - x)\,dx$$

Rule: Instantly eliminates cumbersome denominators by replacing $x$ with $a+b-x$ and adding the original integral to the transformed integral.

Even and Odd Function Parity

$$\int_{-a}^{a} f(x)\,dx = \begin{cases} 2 \int_{0}^{a} f(x)\,dx & \text{if } f(-x) = f(x) \\ 0 & \text{if } f(-x) = -f(x) \end{cases}$$

Rule: Check parity immediately when limits are symmetric $[-a, a]$. If odd, write $0$ instantly without integration.

Wallis's Formula (Reduction for Trig Powers)

$$\int_0^{\pi/2} \sin^n x \cos^m x \, dx = \frac{(n-1)(n-3)\cdots \times (m-1)(m-3)\cdots}{(n+m)(n+m-2)\cdots} \times k$$

Rule: Where $k = \frac{\pi}{2}$ if both $n$ and $m$ are even integers; otherwise $k = 1$.

Application: Standard time-saver for definite integrals of powers of $\sin x$ and $\cos x$ from $0$ to $\frac{\pi}{2}$.

Periodic Property

$$\int_0^{nT} f(x)\,dx = n \int_0^T f(x)\,dx \quad \text{if } f(x+T) = f(x)$$

Rule: Pull out the multiple $n$ if the interval length is an exact integer multiple of the fundamental period $T$.

Application: Frequently tested with fractional parts, modulus of periodic functions, and trigonometric combinations.

Integral Inequality & Estimation

$$\left| \int_a^b f(x)\,dx \right| \le \int_a^b |f(x)| \, dx$$

Rule: Use bounds $m \le f(x) \le M$ to bound the integral by $m(b-a) \le \int_a^b f(x)\,dx \le M(b-a)$ without full evaluation.

Application: Used in option elimination for difficult non-integrable expressions in DSSSB PGT.

Differentiation Under the Integral Sign (Leibniz Rule)

$$\frac{d}{dx} \left[ \int_{u(x)}^{v(x)} f(t)\,dt \right] = f(v(x))v'(x) - f(u(x))u'(x)$$

Rule: Differentiate parameter limits directly without evaluating the integral first.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Ignoring Discontinuities within the Interval

Common Pitfall: Applying fundamental theorem blindly across a point where the integrand is undefined or discontinuous (e.g., $\int_{-1}^{1} \frac{1}{x^2}\,dx$).

Counterexample: $$\int_{-1}^{1} \frac{1}{x^2}\,dx$$

Why it fails: The integrand is discontinuous at $x = 0$. Using $\left[ -\frac{1}{x} \right]_{-1}^1 = -1 - (1) = -2$ is catastrophically false because the integral diverges to $+\infty$.

🛡️ Exam Defense: Always check if the denominator vanishes or if the function has vertical asymptotes inside $[a, b]$. If so, split the integral and test for convergence.

⚠️ COMMON ERROR #02

Modulus Function Blind Splitting

Common Pitfall: Integrating $|f(x)|$ without breaking the interval at the critical points where $f(x) = 0$.

Counterexample: $$\int_{-2}^{2} |x|\,dx$$

Why it fails: Failing to split into $\int_{-2}^{0} (-x)\,dx + \int_{0}^{2} x\,dx$ leads to sign errors.

🛡️ Exam Defense: Always locate roots of the expression inside absolute value symbols within the integration domain before applying integration formulas.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
King's Property	Continuity on $[a, b]$	$$\int_a^b f(x)dx = \int_a^b f(a+b-x)dx$$	Symmetric horizontal shift of area	Sum of lower and upper limits is a clean constant like $\pi/2$ or $1$.
Even/Odd Parity	Symmetric limits $[-a, a]$	$2\int_0^a f(x)dx$ (even) or $0$ (odd)	Cancellation of symmetric signed areas	Limits are $-a$ to $a$ with algebraic/trig terms.
Wallis Formula	Limits $0$ to $\pi/2$, powers of $\sin, \cos$	Product of numerator/denominator factorial-like terms	Area under oscillating trigonometric powers	Integrals of $\sin^n x \cos^m x$ from $0$ to $\pi/2$.
Periodic Property	Function periodic with period $T$	$$\int_0^{nT} f(x)dx = n\int_0^T f(x)dx$$	Repetition of identical area blocks	Limits span multiples of periods like $0$ to $2\pi$ or $0$ to $n\pi$.

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Evaluate the definite integral $I = \int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x} \, dx$.

(A) $\frac{\pi}{4}$(B) $\frac{\pi}{2}$(C) $\pi$(D) $0$

📚 Method A: Formal Textbook Derivation

Let $I = \int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x} \, dx$. Using King's property $\int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx$, we get $I = \int_0^{\pi/2} \frac{\sin(\frac{\pi}{2}-x)}{\sin(\frac{\pi}{2}-x) + \cos(\frac{\pi}{2}-x)} \, dx = \int_0^{\pi/2} \frac{\cos x}{\cos x + \sin x} \, dx$. Adding both equations: $2I = \int_0^{\pi/2} \frac{\sin x + \cos x}{\sin x + \cos x} \, dx = \int_0^{\pi/2} 1\,dx = \frac{\pi}{2}$. Hence, $I = \frac{\pi}{4}$.

⚡ Method B: 25-Second Exam Shortcut

Standard King's property result for $\int_a^b \frac{f(x)}{f(x)+f(a+b-x)}\,dx$ is always $\frac{b-a}{2}$. Here $\frac{\pi/2 - 0}{2} = \frac{\pi}{4}$.

💡 Tip: Recognize this template instantly in UP TGT/PGT to write the answer in 2 seconds.

CANONICAL PROBLEM 02 Dual-Method Breakdown

Evaluate the value of $\int_{-1}^{1} x^5 \cos^4 x \, dx$.

(A) $2$(B) $\frac{\pi}{2}$(C) $0$(D) $1$

📚 Method A: Formal Textbook Derivation

Let $f(x) = x^5 \cos^4 x$. Then $f(-x) = (-x)^5 \cos^4(-x) = -x^5 \cos^4 x = -f(x)$. Thus, the integrand is an odd function. Since the integration limits are symmetric $[-1, 1]$, the integral is identically $0$.

⚡ Method B: 25-Second Exam Shortcut

Check power of $x$: odd power ($5$) multiplied by even power function ($\cos^4 x$) yields an odd function over symmetric limits $\Rightarrow$ Answer is $0$.

💡 Tip: Do not attempt integration by parts. Check parity first whenever limits are symmetric.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: The value of $\int_0^{\pi} x \sin^2 x \, dx$ is:

A $\frac{\pi^2}{4}$

B $\frac{\pi^2}{2}$

C $\frac{\pi}{4}$

D $\pi^2$

Derivation & Analysis:

Using King's property, $I = \int_0^{\pi} (\pi - x)\sin^2(\pi-x)\,dx = \pi \int_0^{\pi} \sin^2 x \, dx - I \implies 2I = \pi \int_0^{\pi} \sin^2 x \, dx = \pi \left(\frac{\pi}{2}\right) = \frac{\pi^2}{2}$. Thus $I = \frac{\pi^2}{4}$.

MCQ 2: Evaluate $\int_{-1}^{1} |x| \, dx$.

A $0$

B $1$

C $2$

D $\frac{1}{2}$

Derivation & Analysis:

Since $|x|$ is an even function on $[-1, 1]$, $\int_{-1}^{1} |x|\,dx = 2 \int_0^1 x\,dx = 2 \left[ \frac{x^2}{2} \right]_0^1 = 1$.

MCQ 3: The value of $\int_0^{\pi/2} \sin^6 x \, dx$ is:

A $\frac{5\pi}{32}$

B $\frac{5\pi}{16}$

C $\frac{3\pi}{16}$

D $\frac{\pi}{32}$

Derivation & Analysis:

By Wallis's formula: $\frac{5 \times 3 \times 1}{6 \times 4 \times 2} \times \frac{\pi}{2} = \frac{15}{48} \times \frac{\pi}{2} = \frac{5}{16} \times \frac{\pi}{2} = \frac{5\pi}{32}$.

MCQ 4: What is the value of $\int_0^{\pi/2} \log(\tan x) \, dx$?

A $\pi$

B $\frac{\pi}{2}$

C $1$

D $0$

Derivation & Analysis:

Let $I = \int_0^{\pi/2} \log(\tan x)\,dx$. Using $\int_0^a f(x)dx = \int_0^a f(a-x)dx$, $I = \int_0^{\pi/2} \log(\cot x)\,dx = -\int_0^{\pi/2} \log(\tan x)\,dx = -I$. Therefore, $2I = 0 \implies I = 0$.

MCQ 5: The value of $\int_0^1 \frac{dx}{1 + x^2}$ is:

A $\frac{\pi}{2}$

B $\frac{\pi}{4}$

C $\frac{\pi}{3}$

D $\frac{\pi}{6}$

Derivation & Analysis:

$$\int_0^1 \frac{dx}{1+x^2} = [\tan^{-1} x]_0^1 = \tan^{-1}(1) - \tan^{-1}(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4}$$.

MCQ 6: If $f(x)$ is a continuous function such that $f(a+b-x) = f(x)$, then $\int_a^b x f(x) \, dx$ is equal to:

A $\frac{a+b}{2} \int_a^b f(x) \, dx$

B $(b-a) \int_a^b f(x) \, dx$

C $0$

D $\int_a^b f(x) \, dx$

Derivation & Analysis:

Standard property of definite integrals: $\int_a^b x f(x)\,dx = \frac{a+b}{2} \int_a^b f(x)\,dx$ whenever $f(a+b-x) = f(x)$.

MCQ 7: The value of $\int_{-a}^{a} \sin^7 x \, dx$ is:

A $2$

B $\pi$

C $0$

D $1$

Derivation & Analysis:

The integrand $\sin^7 x$ is an odd function because $\sin^7(-x) = -\sin^7 x$. Since the interval $[-a, a]$ is symmetric, the integral is $0$.

MCQ 8: Evaluate $\int_0^1 x e^x \, dx$.

A $1$

B $e$

C $0$

D $e-1$

Derivation & Analysis:

Integration by parts: $\int_0^1 x e^x dx = [x e^x]_0^1 - \int_0^1 e^x dx = (1\cdot e^1 - 0) - [e^x]_0^1 = e - (e - 1) = 1$.

MCQ 9: The value of $\int_0^{\pi/2} \frac{\cos x - \sin x}{1 + \sin x \cos x} \, dx$ is:

A $1$

B $\frac{\pi}{2}$

C $\pi$

D $0$

Derivation & Analysis:

Substitute $t = \sin x + \cos x$, or apply symmetry properties observing that the numerator is the derivative of the denominator structure or anti-symmetric under $x \to \pi/2 - x$, evaluating directly to $0$.

Q10

MCQ 10: If $I_n = \int_0^{\pi/2} \sin^n x \, dx$, then the recurrence relation is:

A $I_n = \frac{n-1}{n} I_{n-2}$

B $I_n = \frac{n}{n-1} I_{n-2}$

C $I_n = n I_{n-1}$

D $I_n = \frac{n-1}{n+1} I_{n-2}$

Derivation & Analysis:

Standard reduction formula for powers of sine is $I_n = \frac{n-1}{n} I_{n-2}$.

Common questions

Use King's Property ($\int_a^b f(x)dx = \int_a^b f(a+b-x)dx$) whenever the denominator is symmetric under the transformation $x \to a+b-x$, such as trigonometric sums involving $\sin(x)$ and $\cos(x)$, or logarithmic expressions.

Failing to change the limits of integration when substituting a new variable. Always convert lower and upper limits using $t = g(x)$ rather than substituting back to $x$ at the end.

Memorize standard results like $\int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x}dx = \frac{\pi}{4}$, Wallis formula coefficients, and quick parity checks for symmetric intervals to solve problems.