2. Calculus — Integration

Definite Integral

The definite integral bridges discrete summation and continuous accumulation, representing the net signed area under a curve between explicit limits. This topic covers its fundamental properties, symmetry rules, and limit-sum relations.

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 (First and Second Parts)

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

📐 Geometric Interpretation

Geometrically, the definite integral $\int_a^b f(x)\,dx$ measures the net accumulated area bounded by the curve $y = f(x)$, the x-axis, and the vertical lines $x = a$ and $x = b$, where areas below the x-axis carry a negative sign.

⚙️ Algebraic & Structural Meaning

Algebraically, it establishes the inverse relationship between differentiation and integration, reducing the evaluation of continuous summations to the evaluation of antiderivatives at boundary points.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Continuity on the closed interval	`[a, b] (Closed interval)`	Strictly Mandatory / Necessary	If $f$ has essential or jump discontinuities, Riemann integrability can fail, and standard antiderivative evaluations yield incorrect values.
Integrability / Boundedness	`[a, b] (Closed interval)`	Strictly Mandatory	Unbounded functions on a finite interval fail proper Riemann integration unless treated as improper integrals.
Differentiability of Antiderivative	`(a, b) (Open interval)`	Essential for FTC Part 1	Failure of differentiability breaks the derivative-integral reciprocity, preventing standard application of the Newton-Leibniz formula.

Useful results and shortcuts

Use these identities and results when their conditions apply:

King's Property (Symmetric Limits)

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

Rule: Instantly replaces $x$ with $a+b-x$ to combine denominators or simplify intractable trigonometric/algebraic integrands.

Application: Directly asked in UP PGT/TGT and KVS to evaluate integrals like $\int_0^{\pi/2} \frac{\sin^n x}{\sin^n x + \cos^n x}\,dx = \frac{\pi}{4}$.

Parity Property for Symmetric Intervals

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

Rule: Check if the integrand is even or odd before executing lengthy substitution or integration by parts.

Application: Standard time-saver for polynomials multiplied by trigonometric functions over $[-\pi/2, \pi/2]$ or $[-a, a]$.

Periodic Integration Property

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

Rule: Pull out period multiples to reduce large upper limits to basic periods.

Application: Frequently tested with fractional parts $\{x\}$, $|\sin x|$, or $|\cos x|$ in DSSSB PGT.

Wallis's Formula for Trigonometric Powers

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

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

Application: Essential for rapid calculation of high-power trigonometric definite integrals in UP TGT/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 limits and integrand simultaneously without full evaluation.

Inequality and Estimation Bounds

$$\int_a^b f(x)\%dx \le (b-a) \max_{[a,b]} f(x)$$

Rule: Quickly bound the value of an integral to eliminate incorrect options in multiple-choice questions.

Application: Used for difficult non-integrable functions in advanced PGT screening rounds.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Ignoring Discontinuities Inside the Interval

Common Pitfall: Applying fundamental theorem formulas across points where the integrand or its derivative is undefined (e.g., logarithmic singularities or absolute value vertex).

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

Why it fails: The integrand is discontinuous at $x = 0$, making it an improper integral. Blindly using $[-\frac{1}{x}]_{-1}^1$ yields $-2$, which is absurd since $\frac{1}{x^2} > 0$.

🛡️ Exam Defense: Always scan the integration interval $[a, b]$ for points where the denominator vanishes or the function definition changes.

⚠️ COMMON ERROR #02

Misapplication of Periodic or Symmetric Properties

Common Pitfall: Using odd/even function shortcuts when the interval is not symmetric about the origin (e.g., from $0$ to $a$ instead of $-a$ to $a$).

Counterexample: $$\int_{0}^{\pi} \sin x \,dx$$

Why it fails: Although $\sin x$ is odd about $x = 0$, the interval is not symmetric around $0$, so setting the integral to zero is completely false.

🛡️ Exam Defense: Verify that the lower limit is the exact negative of the upper limit before invoking parity properties.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
King's Property	Continuity on $[a, b]$	$\int_a^b f(x)dx = \int_a^b f(a+b-x)dx$	Horizontal reflection and vertical translation symmetry	Denominators like $f(x) + f(a+b-x)$
Parity Property	Continuity on $[-a, a]$	Symmetric bounds reduction	Cancellation of symmetric areas	Limits of type $[-a, a]$ with odd/even powers
Wallis Formula	Continuous powers of sine/cosine	Product of factorial-like reductions	Area under trigonometric polynomial curves	Limits from $0$ to $\pi/2$ with powers of $\sin$ and $\cos$
Leibniz Integral Rule	Continuity and differentiability of limits	$\frac{d}{dx}\int_{u(x)}^{v(x)} f(t)dt$	Rate of change of accumulated area with moving boundaries	Integral equated to a function with variable limits

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

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

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

📚 Method A: Formal Textbook Derivation

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

⚡ Method B: 25-Second Exam Shortcut

Standard King's property result for $\frac{f(x)}{f(x)+f(a+b-x)}$ always evaluates to $\frac{b-a}{2} = \frac{\pi/2 - 0}{2} = \frac{\pi}{4}$.

💡 Tip: Recognize this standard pattern instantly to save 2 minutes of calculation.

CANONICAL PROBLEM 02 Dual-Method Breakdown

Evaluate the value of: $$\int_{-1}^{1} \log\left(\frac{2-x}{2+x}\right)\,dx$$

(A) $\log 2$(B) $0$(C) $1$(D) $-\log 2$

📚 Method A: Formal Textbook Derivation

Let $f(x) = \log\left(\frac{2-x}{2+x}\right)$. Then $f(-x) = \log\left(\frac{2+x}{2-x}\right) = -\log\left(\frac{2-x}{2+x}\right) = -f(x)$. Since the integrand is an odd function and the interval is $[-1, 1]$, the integral is strictly $0$.

⚡ Method B: 25-Second Exam Shortcut

Check symmetry $f(-x) = -f(x)$ over symmetric interval $[-a, a]$ -> Output is immediately $0$.

💡 Tip: Look for logarithmic expressions with $(a-x)/(a+x)$ over symmetric limits.

Practice questions

Choose an answer to check it and see the solution:

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: $I = \int_0^\pi (\pi-x)\sin^2(\pi-x)dx = \pi \int_0^\pi \sin^2 x dx - I \implies 2I = \pi \int_0^\pi \frac{1-\cos 2x}{2} dx = \pi \cdot \frac{\pi}{2} = \frac{\pi^2}{2}$, so $I = \frac{\pi^2}{4}$.

If $f(x)$ is an even function, then $\int_{-a}^{a} f(x)\,dx$ is equal to:

A 0

B $2 \int_0^a f(x)\,dx$

C $\int_0$^a f(x)\,dx

D $a \int_0^a f(x)\,dx$

Derivation & Analysis:

By definition of even function symmetry, area from $-a$ to $0$ equals area from $0$ to $a$, doubling the integral over $[0, a]$. fielder

The value of $\int_0^{\pi/2} \sin^5 x \cos^4 x \,dx$ is:

A $\frac{8}{315}$

B $\frac{4}{315}$

C $\frac{16}{315}$

D $\frac{1}{63}$

Derivation & Analysis:

Using Wallis's formula with $n=5$ (odd) and $m=4$ (even): $\frac{(5-1)(3)}{(5+4)(5+4-2)(5+4-4)(5+4-6)} \times \text{term} = \frac{4 \times 2 \times 3 \times 1}{9 \times 7 \times 5 \times 3} = \frac{48}{945} = \frac{8}{315}$.

What is the derivative of $\int_{0}^{x^2} \cos(t^2)\,dt$ with respect to $x$?

A $2x \cos(x^4)$

B $\cos(x^2)$

C $2x \cos(x^2)$

D $\cos(x^4)$

Derivation & Analysis:

By Leibniz integral rule: $\frac{d}{dx} \int_{0}^{x^2} \cos(t^2)dt = \cos((x^2)^2) \cdot \frac{d}{dx}(x^2) - 0 = \cos(x^4) \cdot 2x = 2x \cos(x^4)$.

The value of $\int_{-2}^{2} |x| \,dx$ is:

A 0

B 2

C 4

D 8

Derivation & Analysis:

$|x|$ is even, so $\int_{-2}^2 |x|dx = 2 \int_0^2 x dx = 2 \left[ \frac{x^2}{2} \right]_0^2 = 2(2) = 4$.

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

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

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

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

D $\pi$

Derivation & Analysis:

Standard antiderivative: $\int \frac{dx}{1+x^2} = \tan^{-1}(x)$. Evaluating from $0$ to $1$ gives $\tan^{-1}(1) - \tan^{-1}(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4}$.

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

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

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

C $\frac{\pi^2}{8}$

D $\pi^2$

Derivation & Analysis:

Using King's property, $I = \int_0^\pi \frac{(\pi-x)\sin(\pi-x)}{1+\cos^2(\pi-x)}dx = \pi \int_0^\pi \frac{\sin x}{1+\cos^2 x}dx - I \implies 2I = \pi \int_0^\pi \frac{\sin x}{1+\cos^2 x}dx$. Substituting $\cos x = t$ leads to $\frac{\pi^2}{4}$, so $I = \frac{\pi^2}{8}$.

If $f(x)$ is continuous on $[a, b]$, then the average value of $f(x)$ on $[a, b]$ is given by:

A $\frac{1}{b-a} \int_a^b f(x)\,dx$

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

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

D $\frac{1}{b} \int_a^b f(x)\,dx$

Derivation & Analysis:

The mean value theorem for integrals defines the average value as $\frac{1}{b-a} \int_a^b f(x)\,dx$.

The value of $\int_{-1}^{1} x^5 \cos^4 x \,dx$ is:

A $\frac{1}{2}$

B 1

C 0

D $\frac{\pi}{2}$

Derivation & Analysis:

The integrand $x^5 \cos^4 x$ is an odd function because $(-x)^5 \cos^4(-x) = -x^5 \cos^4 x$. Over the symmetric interval $[-1, 1]$, the integral of an odd function is strictly $0$.

Q10

Evaluate: $\int_0^1 x e^x \,dx$

A 1

B e

C 0

D 2e - 1

Derivation & Analysis:

Using integration by parts: $\int x e^x dx = x e^x - \int e^x dx = x e^x - e^x = (x-1)e^x$. Evaluating from $0$ to $1$: $(0)e^1 - (-1)e^0 = 0 - (-1) = 1$.

Common questions

Use King's Property ($\int_a^b f(x)dx = \int_a^b f(a+b-x)dx$) whenever the denominator is invariant under reflection or when trigonometric limits are symmetric around $\pi/2$ or $a+b-x$ simplifies the integrand.

Forgetting to check for infinite discontinuities or sign changes inside the integration interval (such as modulus functions or logarithmic denominators vanishing at interior points).

Memorize standard definite integral results (like Wallis formulas and King's property standard forms) and always check for even/odd parity on symmetric limits before attempting full expansion.

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