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Class 11 ? 12 Bridge � Integral Calculus � JEE Masterclass

Integral Calculus � Complete JEE Module

The Core Mission

Integration is the reverse of differentiation. If differentiation splits a function into its rate of change, integration re-assembles those pieces to find the original function, an area, a volume, or an accumulated quantity. For JEE, integration constitutes approximately 30�35% of the Mathematics paper. This masterclass covers every technique you need � no external notes required.

PART 1: INDEFINITE INTEGRATION

1. What is an Integral?

If \(\frac{d}{dx}[F(x)] = f(x)\), then \(\int f(x)\,dx = F(x) + C\), where \(C\) is the constant of integration. This \(C\) represents the family of all anti-derivatives. It is non-negotiable � always write it.

Error Trap Missing + C is a guaranteed mark loss in JEE. The integral of a function is a family of curves, not a single curve. Every indefinite integral answer must end with \(+ C\).

2. Standard Integral Formulas (Must Memorise)

Function \(f(x)\) \(\int f(x)\,dx\)
\(x^n\) \((n \ne -1)\) \(\dfrac{x^{n+1}}{n+1} + C\)
\(\dfrac{1}{x}\) \(\ln|x| + C\)
\(e^x\) \(e^x + C\)
\(a^x\) \(\dfrac{a^x}{\ln a} + C\)
\(\sin x\) \(-\cos x + C\)
\(\cos x\) \(\sin x + C\)
\(\tan x\) \(\ln|\sec x| + C\)
\(\cot x\) \(\ln|\sin x| + C\)
\(\sec x\) \(\ln|\sec x + \tan x| + C\)
\(\csc x\) \(\ln|\csc x - \cot x| + C\)
\(\sec^2 x\) \(\tan x + C\)
\(\csc^2 x\) \(-\cot x + C\)
\(\sec x \tan x\) \(\sec x + C\)
\(\csc x \cot x\) \(-\csc x + C\)
Function \(f(x)\) \(\int f(x)\,dx\)
\(\dfrac{1}{\sqrt{1-x^2}}\) \(\sin^{-1}x + C\)
\(\dfrac{-1}{\sqrt{1-x^2}}\) \(\cos^{-1}x + C\)
\(\dfrac{1}{1+x^2}\) \(\tan^{-1}x + C\)
\(\dfrac{1}{a^2+x^2}\) \(\dfrac{1}{a}\tan^{-1}\!\dfrac{x}{a} + C\)
\(\dfrac{1}{\sqrt{a^2-x^2}}\) \(\sin^{-1}\!\dfrac{x}{a} + C\)
\(\dfrac{1}{x^2-a^2}\) \(\dfrac{1}{2a}\ln\!\left|\dfrac{x-a}{x+a}\right| + C\)
\(\dfrac{1}{a^2-x^2}\) \(\dfrac{1}{2a}\ln\!\left|\dfrac{a+x}{a-x}\right| + C\)
\(\dfrac{1}{\sqrt{x^2+a^2}}\) \(\ln\!\left|x+\sqrt{x^2+a^2}\right| + C\)
\(\dfrac{1}{\sqrt{x^2-a^2}}\) \(\ln\!\left|x+\sqrt{x^2-a^2}\right| + C\)
\(\sqrt{a^2-x^2}\) \(\dfrac{x}{2}\sqrt{a^2-x^2}+\dfrac{a^2}{2}\sin^{-1}\!\dfrac{x}{a}+C\)
\(\sqrt{x^2 \pm a^2}\) \(\dfrac{x}{2}\sqrt{x^2 \pm a^2} \pm \dfrac{a^2}{2}\ln\!\left|x+\sqrt{x^2 \pm a^2}\right|+C\)

3. Method 1 � Integration by Substitution

Core Idea: Replace a complicated part of the integrand with a single variable \(t\) to simplify it. If you can spot \(f'(x)\) sitting alongside \(f(x)\), substitute \(u = f(x)\).

Golden Rule

If the integrand has \([f(x)]^n \cdot f'(x)\) form: Put \(t = f(x)\), so \(dt = f'(x)\,dx\). The integral becomes \(\int t^n\,dt = \dfrac{t^{n+1}}{n+1} + C\).

Q1. Evaluate \(\displaystyle\int \frac{2x}{x^2+1}\,dx\) Solution: Let \(t = x^2 + 1\), so \(dt = 2x\,dx\).
\(\displaystyle\int \frac{dt}{t} = \ln|t| + C = \boxed{\ln(x^2+1) + C}\)
Q2. Evaluate \(\displaystyle\int \frac{\sin(\ln x)}{x}\,dx\) Solution: Let \(t = \ln x\), so \(dt = \dfrac{1}{x}\,dx\).
\(\displaystyle\int \sin(t)\,dt = -\cos t + C = \boxed{-\cos(\ln x) + C}\)
Q3. Evaluate \(\displaystyle\int \tan x\,dx\) Solution: Write \(\tan x = \dfrac{\sin x}{\cos x}\). Let \(t = \cos x\), so \(dt = -\sin x\,dx\).
\(\displaystyle\int \frac{-dt}{t} = -\ln|t| + C = \boxed{\ln|\sec x| + C}\)

4. Method 2 � Integration by Parts (IBP)

Used when the integrand is a product of two different types of functions (e.g., polynomial � trig, polynomial � log, exponential � trig).

Formula \[\int u \cdot v\,dx = u \int v\,dx - \int \left(\frac{du}{dx} \cdot \int v\,dx\right)dx\]

ILATE Rule � Choose \(u\) (to differentiate) in this priority order:
Inverse Trig ? Logarithmic ? Algebraic (polynomial) ? Trigonometric ? Exponential

Q4. Evaluate \(\displaystyle\int x \cos x\,dx\) Solution: Let \(u = x\) (A from ILATE), \(v = \cos x\).
\[\int x\cos x\,dx = x(\sin x) - \int 1 \cdot \sin x\,dx = x\sin x + \cos x + C\] \[\boxed{= x\sin x + \cos x + C}\]
Q5. Evaluate \(\displaystyle\int x^2 e^x\,dx\) Solution: Apply IBP twice. Let \(u=x^2\), \(v=e^x\).
\(= x^2 e^x - \int 2x e^x\,dx\)
Apply IBP again on \(\int 2x e^x\,dx\): \(= 2xe^x - 2e^x\)
\[\boxed{\int x^2 e^x\,dx = e^x(x^2 - 2x + 2) + C}\]
Q6. JEE Key Form � Evaluate \(\displaystyle\int e^x[\sin x + \cos x]\,dx\) Solution: This matches the JEE standard formula: \(\int e^x[f(x)+f'(x)]\,dx = e^x f(x) + C\).
Here \(f(x)=\sin x\), \(f'(x)=\cos x\).
\[\boxed{= e^x \sin x + C}\]
JEE Standard Form � Memorise \[\int e^x[f(x) + f'(x)]\,dx = e^x f(x) + C\] This pattern appears almost every year in JEE Main. Identify \(f(x)\) and verify its derivative is the other bracket term.

5. Method 3 � Partial Fractions

Used for integrating rational functions \(\dfrac{P(x)}{Q(x)}\) where degree of \(P <\) degree of \(Q\), and \(Q(x)\) can be factored.

Type of Factor in Denominator Partial Fraction Form
Linear: \((x-a)\) \(\dfrac{A}{x-a}\)
Repeated linear: \((x-a)^2\) \(\dfrac{A}{x-a} + \dfrac{B}{(x-a)^2}\)
Irreducible quadratic: \((x^2+bx+c)\) \(\dfrac{Ax+B}{x^2+bx+c}\)
Q7. Evaluate \(\displaystyle\int \frac{1}{(x-1)(x+2)}\,dx\) Solution: Decompose: \(\dfrac{1}{(x-1)(x+2)} = \dfrac{A}{x-1}+\dfrac{B}{x+2}\).
Multiply through: \(1 = A(x+2)+B(x-1)\).
Put \(x=1\): \(1=3A \Rightarrow A=1/3\). Put \(x=-2\): \(1=-3B \Rightarrow B=-1/3\).
\[\int \frac{1}{3}\cdot\frac{1}{x-1}\,dx - \int\frac{1}{3}\cdot\frac{1}{x+2}\,dx = \boxed{\frac{1}{3}\ln\left|\frac{x-1}{x+2}\right| + C}\]

6. Special Trigonometric Integrals

Key Reduction Formulas
  • \(\sin^2 x = \dfrac{1-\cos 2x}{2}\)    \(\cos^2 x = \dfrac{1+\cos 2x}{2}\)
  • \(\sin^3 x = \dfrac{3\sin x - \sin 3x}{4}\)    \(\cos^3 x = \dfrac{3\cos x + \cos 3x}{4}\)
  • \(\int \sin^m x \cos^n x\,dx\): Use Walli's formula for definite version. For indefinite, use half-angle or reduction.
Q8. Evaluate \(\displaystyle\int \sin^2 x\,dx\) Solution: \(\sin^2 x = \dfrac{1-\cos 2x}{2}\)
\(\displaystyle\int \frac{1-\cos 2x}{2}\,dx = \frac{x}{2} - \frac{\sin 2x}{4} + C = \boxed{\frac{x}{2} - \frac{\sin 2x}{4} + C}\)
Q9. Evaluate \(\displaystyle\int \frac{1}{a+b\cos x}\,dx\) using the Weierstrass substitution Solution � The T-Substitution \(\left(t = \tan\dfrac{x}{2}\right)\): Recall: \(\cos x = \dfrac{1-t^2}{1+t^2}\), \(\sin x = \dfrac{2t}{1+t^2}\), \(dx = \dfrac{2\,dt}{1+t^2}\).
\[\int \frac{1}{a+b\cdot\frac{1-t^2}{1+t^2}}\cdot\frac{2\,dt}{1+t^2} = \int \frac{2\,dt}{(a+b)+(a-b)t^2}\] This reduces to \(\dfrac{1}{a^2-b^2}\)-type standard form. This substitution works for any rational function of \(\sin x\) and \(\cos x\).

7. Method 4 � Integration of the form \(\int \frac{px+q}{ax^2+bx+c}\,dx\) and \(\int \frac{px+q}{\sqrt{ax^2+bx+c}}\,dx\)

Split the numerator \(px+q\) as \(\lambda(2ax+b) + \mu\). Find \(\lambda\) and \(\mu\) by comparing coefficients. This decomposes the integral into a log part and a standard inverse-trig or log part.

Q10. Evaluate \(\displaystyle\int \frac{x+3}{x^2+2x+5}\,dx\) Solution: Write \(x+3 = \lambda(2x+2)+\mu\).
Comparing: \(\lambda=\frac12\), \(\mu=2\).
\[= \frac{1}{2}\int\frac{2x+2}{x^2+2x+5}\,dx + 2\int\frac{dx}{(x+1)^2+4}\] \[= \frac{1}{2}\ln(x^2+2x+5) + \tan^{-1}\!\frac{x+1}{2} + C\] \[\boxed{= \frac{1}{2}\ln(x^2+2x+5) + \tan^{-1}\!\frac{x+1}{2} + C}\]

PART 2: DEFINITE INTEGRATION

8. The Fundamental Theorem of Calculus

If \(F(x)\) is the anti-derivative of \(f(x)\), then:

Core Formula \[\int_a^b f(x)\,dx = \Big[F(x)\Big]_a^b = F(b) - F(a)\]

The definite integral gives a specific numerical value (no + C), representing the net signed area under the curve from \(x=a\) to \(x=b\).

9. Properties of Definite Integrals (JEE Examiner's Favourites)

#PropertyDescription
P1\(\int_a^b f\,dx = -\int_b^a f\,dx\)Swapping limits flips sign.
P2\(\int_a^b f\,dx = \int_a^c f\,dx + \int_c^b f\,dx\)Breaking at any interior point.
P3\(\int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx\)King's Rule — Most Powerful JEE Tool.
P4\(\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx\)Special King's when \(b=a\).
P5\(\int_{-a}^a f(x)\,dx = 2\int_0^a f(x)\,dx\) if \(f\) is evenSymmetry about y-axis.
P6\(\int_{-a}^a f(x)\,dx = 0\) if \(f\) is oddAnti-symmetry about origin.
P7\(\int_0^{2a} f(x)\,dx = 2\int_0^a f(x)\,dx\) if \(f(2a-x)=f(x)\)If \(f(2a-x)=-f(x)\), integral = 0.
P8\(\int_0^{nT} f\,dx = n\int_0^{T} f\,dx\)Periodicity Property (\(T\) = period).
Q11. JEE Classic — Evaluate \(\displaystyle I = \int_0^{\pi} \frac{x\sin x}{1+\cos^2 x}\,dx\) Solution — Using King's Property (P3): Apply \(x \to (\pi - x)\): \(I = \displaystyle\int_0^{\pi} \frac{(\pi-x)\sin x}{1+\cos^2 x}\,dx\)

Adding the two expressions: \(2I = \pi\displaystyle\int_0^{\pi}\frac{\sin x}{1+\cos^2 x}\,dx\)

Let \(t = \cos x\), \(dt = -\sin x\,dx\). Limits: \(t: 1 \to -1\).
\(2I = -\pi\displaystyle\int_1^{-1}\frac{dt}{1+t^2} = \pi\Big[\tan^{-1}t\Big]_{-1}^{1} = \pi\cdot\frac{\pi}{2} = \frac{\pi^2}{2}\)
\(\boxed{I = \dfrac{\pi^2}{4}}\)
Q12. Evaluate \(\displaystyle\int_{-\pi}^{\pi} (x^3 + \sin^3 x)\,dx\) Solution — Odd Function Property (P6): Both \(x^3\) and \(\sin^3 x\) are odd functions. For any odd \(f\) over \([-a,a]\): integral = 0.
\(\boxed{= 0}\). No computation required!

10. Walli's Formula

Formula \[\int_0^{\pi/2} \sin^m x \cos^n x\,dx = \frac{(m-1)!!(n-1)!!}{(m+n)!!} \times K\] where \(K = \dfrac{\pi}{2}\) if both \(m\) and \(n\) are even integers; \(K = 1\) otherwise.
Note: \(\displaystyle\int_0^{\pi/2}\sin^n x\,dx = \int_0^{\pi/2}\cos^n x\,dx\) always (by King's Rule).
Q13. Evaluate \(\displaystyle\int_0^{\pi/2} \sin^4 x\,dx\) Solution: \(m=4\) (even), \(n=0\) (treat as even). \(K = \pi/2\).
\(= \dfrac{(3)(1)}{(4)(2)} \cdot \dfrac{\pi}{2} = \dfrac{3}{8}\cdot\dfrac{\pi}{2} = \boxed{\dfrac{3\pi}{16}}\)

11. Leibniz Rule — Differentiating Under the Integral Sign

Formula \[\frac{d}{dt}\int_{g(t)}^{h(t)} f(x)\,dx = f(h(t))\cdot h'(t) - f(g(t))\cdot g'(t)\]
Q14. If \(F(x) = \displaystyle\int_x^{x^2} \frac{\ln t}{t}\,dt\), find \(F'(2)\) Solution: \(F'(x) = \dfrac{\ln(x^2)}{x^2}\cdot(2x) - \dfrac{\ln x}{x}\cdot 1 = \dfrac{2\ln x}{x} - \dfrac{\ln x}{x} = \dfrac{\ln x}{x}\)
At \(x=2\): \(\boxed{F'(2) = \dfrac{\ln 2}{2}}\)

12. Limit as a Definite Integral (JEE Regular)

Formula \[\lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{n} f\!\left(\frac{r}{n}\right) = \int_0^1 f(x)\,dx\] Or more generally: \(\displaystyle\lim_{n\to\infty}\frac{1}{n}\sum_{r=p}^{qn}f\!\left(\frac{r}{n}\right) = \int_p^q f(x)\,dx\)
Q15. Evaluate \(\displaystyle\lim_{n\to\infty}\left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{3n}\right)\) Solution: \(= \displaystyle\lim_{n\to\infty}\sum_{r=1}^{2n}\frac{1}{n+r} = \lim_{n\to\infty}\frac{1}{n}\sum_{r=1}^{2n}\frac{1}{1+r/n} = \int_0^2 \frac{dx}{1+x} = \Big[\ln(1+x)\Big]_0^2 = \boxed{\ln 3}\)

PART 3: AREA UNDER CURVES (AOI)

13. Core Area Formulas

Area between curve and x-axis:

\(A = \displaystyle\int_a^b |f(x)|\,dx\)

⚠️ Use absolute value! Area is always positive. Split the integral at zeroes of \(f(x)\) if it changes sign.
Area between two curves:

\(A = \displaystyle\int_a^b [f_{\text{upper}}(x) - f_{\text{lower}}(x)]\,dx\)

Limits of integration = x-coordinates of intersection points.
Q16. Area enclosed by \(y = x^2\) and \(y = x\) Solution: Intersection: \(x^2 = x \Rightarrow x=0, 1\). Upper curve in \([0,1]\): \(y=x\).
\(A = \displaystyle\int_0^1 (x-x^2)\,dx = \left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1 = \frac{1}{2}-\frac{1}{3} = \boxed{\frac{1}{6}}\text{ sq. units}\)
Q17. Area bounded by \(y^2=4x\) and \(x^2=4y\) Solution: Intersection: sub \(y=\frac{x^2}{4}\) into \(y^2=4x\): \(\frac{x^4}{16}=4x \Rightarrow x=0,4\).
Both lie on: parabola \(y=2\sqrt{x}\) (above) vs parabola \(y=\frac{x^2}{4}\) (below) in \([0,4]\).
\(A = \displaystyle\int_0^4\!\!\left[2\sqrt{x}-\frac{x^2}{4}\right]dx = \left[\frac{4x^{3/2}}{3}-\frac{x^3}{12}\right]_0^4 = \frac{32}{3}-\frac{16}{3} = \boxed{\frac{16}{3}}\text{ sq. units}\)
Q18. Area between \(y=\sin x\) and \(y=\cos x\) from \(x=\pi/4\) to \(x=5\pi/4\) Solution: In this interval, \(\sin x \ge \cos x\).
\(A = \displaystyle\int_{\pi/4}^{5\pi/4}(\sin x-\cos x)\,dx = [-\cos x-\sin x]_{\pi/4}^{5\pi/4}\)
\(= \left[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right]-\left[-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right] = \sqrt{2}+\sqrt{2} = \boxed{2\sqrt{2}}\text{ sq. units}\)
Q19. Area inside \(|x|+|y|=1\) Solution: The shape is a square (rotated 45°) with vertices at \((\pm1,0)\) and \((0,\pm1)\).
One quarter (first quadrant): \(\displaystyle\int_0^1(1-x)\,dx = \frac{1}{2}\). Total = \(4 \times \frac{1}{2} = \boxed{2}\text{ sq. units}\)

14. Quick Reference: Areas of Standard Curves

CurveArea Formula
Parabola \(y^2=4ax\) cut by \(x=h\)\(\dfrac{2}{3}h \cdot 2\sqrt{ah}\) = \(\frac{2}{3}\) of bounding rectangle
Ellipse \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\)\(\pi ab\)
\(y=\sin x\), one arch \([0,\pi]\)2 sq. units
\(y=|\sin x|\), \([0, n\pi]\)\(2n\) sq. units
Circle \(x^2+y^2=a^2\)\(\pi a^2\)

PART 4: Master Practice Matrix — 20 Extra JEE-Level Solved Problems

Section A: More Indefinite Integration
  1. \(\displaystyle\int \frac{dx}{\sin x \cos^3 x}\)
    Sol: Divide num & denom by \(\cos^4 x\): \(\displaystyle\int\frac{\sec^4 x}{\tan x}\,dx = \int\frac{\sec^2 x(1+\tan^2 x)}{\tan x}\,dx\). Let \(t=\tan x\): \(\displaystyle\int\frac{1+t^2}{t}\,dt = \ln t + \frac{t^2}{2}+C = \ln|\tan x|+\frac{\tan^2 x}{2}+C\)
  2. \(\displaystyle\int \frac{x\,dx}{\sqrt{x^2+a^2}}\)
    Sol: Let \(t=x^2+a^2\), \(dt=2x\,dx\). \(= \frac{1}{2}\displaystyle\int\frac{dt}{\sqrt{t}} = \sqrt{t}+C = \sqrt{x^2+a^2}+C\)
  3. \(\displaystyle\int \cos^4 x\,dx\)
    Sol: \(\cos^4 x = \left(\dfrac{1+\cos 2x}{2}\right)^2 = \dfrac{1+2\cos 2x+\cos^2 2x}{4} = \dfrac{1+2\cos 2x+\frac{1+\cos 4x}{2}}{4}\)
    \(= \dfrac{3}{8}+\dfrac{\cos 2x}{2}+\dfrac{\cos 4x}{8}\). Integrating: \(\dfrac{3x}{8}+\dfrac{\sin 2x}{4}+\dfrac{\sin 4x}{32}+C\)
  4. \(\displaystyle\int \frac{3x+5}{x^3-x^2-x+1}\,dx\)
    Sol: Factor denom: \((x-1)^2(x+1)\). Partial fractions: \(\dfrac{A}{x-1}+\dfrac{B}{(x-1)^2}+\dfrac{C}{x+1}\). Evaluating: \(A=-1,B=4,C=1\). Answer: \(-\ln|x-1|-\dfrac{4}{x-1}+\ln|x+1|+C\).
  5. \(\displaystyle\int \sin^{-1}\!\sqrt{x}\,dx\)
    Sol: Let \(x=\sin^2\theta\), \(dx=2\sin\theta\cos\theta\,d\theta\). \(= \displaystyle\int\theta\cdot 2\sin\theta\cos\theta\,d\theta = \int\theta\sin 2\theta\,d\theta\). IBP: \(= -\dfrac{\theta\cos 2\theta}{2}+\dfrac{\sin 2\theta}{4}+C = x\sin^{-1}\!\sqrt{x}-\frac{1}{2}\sqrt{x(1-x)}\cdot 2+C\)... substituting back: \(\boxed{x\sin^{-1}\!\sqrt{x}+\sqrt{x-x^2}+C}\)
Section B: More Definite Integration
  1. \(\displaystyle\int_0^{\pi/2}\frac{dx}{1+\tan^5 x}\)
    Sol: King's Rule: \(I^* = \displaystyle\int_0^{\pi/2}\frac{1}{1+\cot^5 x}dx\). \(I+I^* = \displaystyle\int_0^{\pi/2}1\,dx = \frac{\pi}{2}\). So \(\boxed{I=\frac{\pi}{4}}\). (Works for any power!)
  2. \(\displaystyle\int_0^{4} \lfloor x \rfloor\,dx\) where \(\lfloor x \rfloor\) = Greatest Integer Function
    Sol: \(= \displaystyle\int_0^1 0\,dx+\int_1^2 1\,dx+\int_2^3 2\,dx+\int_3^4 3\,dx = 0+1+2+3 = \boxed{6}\)
  3. \(\displaystyle\int_0^1 \frac{\ln(1+x)}{1+x^2}\,dx\) (JEE Advanced 2014)
    Sol: Let \(x=\tan\theta\), limits \(0\to\pi/4\). \(= \displaystyle\int_0^{\pi/4}\ln(1+\tan\theta)\,d\theta\). By King's on \([0,\pi/4]\): \(I^* = \int_0^{\pi/4}\ln\!\left(1+\tan\!\left(\frac{\pi}{4}-\theta\right)\!\right)d\theta\). Using \(\tan(\pi/4-\theta)=\frac{1-\tan\theta}{1+\tan\theta}\): \(I^* = \int_0^{\pi/4}\ln\frac{2}{1+\tan\theta}\,d\theta\).
    \(2I = \int_0^{\pi/4}\ln 2\,d\theta = \frac{\pi\ln 2}{4}\). \(\boxed{I = \frac{\pi\ln 2}{8}}\)
  4. \(\displaystyle\lim_{n\to\infty}\frac{1^n+2^n+\cdots+n^n}{n^{n+1}}\)
    Sol: \(= \displaystyle\lim_{n\to\infty}\frac{1}{n}\sum_{r=1}^n\left(\frac{r}{n}\right)^n = \int_0^1 x^{?}\,dx\)... Actually: \(\displaystyle\int_0^1 \lim_{n\to\infty}\left(r/n\right)^n\,dr\). Recognise that \(\left(\frac{r}{n}\right)^n \to 0\) for \(r < n\) but the boundary term gives: using limit-sum formula with \(f(x)=x^n\) is non-trivial. The rigorous answer: \(\frac{1}{e-1}\)... Via a different approach: define \(a_n = \sum_{k=1}^n (k/n)^n/n\). This converges to \(\boxed{\frac{1}{e-1}}\) (famous result).
  5. \(\displaystyle\int_0^{\infty} \frac{dx}{(1+x)(1+x^2)}\)
    Sol: Partial fractions: \(\dfrac{1}{(1+x)(1+x^2)} = \dfrac{A}{1+x}+\dfrac{Bx+C}{1+x^2}\). Solving: \(A=\frac12, B=-\frac12, C=\frac12\).
    \(\displaystyle\int_0^{\infty}\!\!\left[\frac{1/2}{1+x}-\frac{x/2}{1+x^2}+\frac{1/2}{1+x^2}\right]dx\). Each term integrates to give \(\Big[\frac{\ln(1+x)}{2}-\frac{\ln(1+x^2)}{4}+\frac{\tan^{-1}x}{2}\Big]_0^\infty\). Result: \(\frac{\pi}{4}\).
Section C: Area Under Curves
  1. Area bounded by \(y = x^2 - 4\) and \(y = -x^2 + 4\)
    Sol: Intersection: \(x^2-4 = -x^2+4 \Rightarrow 2x^2=8 \Rightarrow x=\pm2\). Upper: \(-x^2+4\).
    \(A = \displaystyle\int_{-2}^2 [(-x^2+4)-(x^2-4)]\,dx = \int_{-2}^2(8-2x^2)\,dx = \left[8x-\frac{2x^3}{3}\right]_{-2}^2 = \frac{64}{3}\) sq. units.
  2. Area bounded by \(y = e^x\), \(y = e^{-x}\) and \(x=1\)
    Sol: \(e^x = e^{-x}\Rightarrow x=0\). On \([0,1]\): \(e^x > e^{-x}\).
    \(A = \displaystyle\int_0^1(e^x-e^{-x})\,dx = [e^x+e^{-x}]_0^1 = (e+e^{-1})-2 = e+\frac{1}{e}-2\) sq. units.
  3. Area of smaller region bounded by \(x^2+y^2=9\) and \(x+y=3\)
    Sol: Circle radius 3, line \(x+y=3\) cuts it. Intersection: \(x=0,y=3\) and \(x=3,y=0\).
    \(A_{\text{seg}} = A_{\text{quarter circle}} - A_{\text{triangle}} = \frac{1}{4}\pi(9)-\frac{1}{2}(3)(3) = \frac{9\pi}{4}-\frac{9}{2}\) sq. units.

PART 5: Critical Results, Error Traps & JEE Shortcuts

Error Trap 1 Absolute Value in Area: \(\displaystyle\int_a^b f(x)\,dx\) gives signed area. If \(f(x) < 0\) in part of the interval, always split at x-intercepts and use \(|f(x)|\) for geometric area.
Error Trap 2 Odd integrands over symmetric limits: \(\displaystyle\int_{-a}^{a}\) of any odd function is always 0 — identify the parity first before computing.
Error Trap 3 Discontinuous functions: If \(f(x)\) has a discontinuity at \(x=c\) inside \([a,b]\), you MUST split: \(\displaystyle\int_a^c + \int_c^b\). Ignoring this leads to completely wrong answers.
Error Trap 4 Area is always positive! If the curve dips below the x-axis, the definite integral gives a negative value, but the area is its absolute value. Many students write negative areas — it is wrong.
JEE Champion Shortcuts Results you MUST memorise:
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