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Class 12 Mathematics · NCERT

Chapter 7: Integrals

Practice Worksheet — Indefinite & Definite Integrals, Methods of Integration, Properties

📝 25 Questions
🔢 MCQ + Numerical
💡 Step-by-step Solutions
🎯 AAI ATC Focused
📋

Section A — Multiple Choice Questions

Choose the most correct option for each question. (Q1–Q15)

1
MCQ · Numerical
The value of \( \displaystyle\int x^3\, dx \) is:
A \( \dfrac{x^4}{4} + C \)
B \( \dfrac{x^3}{3} + C \)
C \( 3x^2 + C \)
D \( x^4 + C \)
✅ Solution Using the power rule: \(\displaystyle\int x^n\,dx = \frac{x^{n+1}}{n+1} + C\) for \(n \neq -1\).
Here \(n = 3\), so \(\displaystyle\int x^3\,dx = \frac{x^4}{4} + C\).
Answer: (A) \( \dfrac{x^4}{4} + C \)
2
MCQ · Numerical
The anti-derivative of \( \cos x \) is:
A \( -\sin x + C \)
B \( \sin x + C \)
C \( \cos x + C \)
D \( \tan x + C \)
✅ Solution We know that \(\dfrac{d}{dx}(\sin x) = \cos x\). Therefore, \(\displaystyle\int \cos x\,dx = \sin x + C\).
Answer: (B) \( \sin x + C \)
3
MCQ · Numerical
Evaluate: \( \displaystyle\int \sec^2 x\, dx \)
A \( \cot x + C \)
B \( -\cot x + C \)
C \( \tan x + C \)
D \( \sec x + C \)
✅ Solution Since \(\dfrac{d}{dx}(\tan x) = \sec^2 x\), we get: \(\displaystyle\int \sec^2 x\,dx = \tan x + C\).
Answer: (C) \( \tan x + C \)
4
MCQ · Numerical
The value of \( \displaystyle\int e^x\, dx \) is:
A \( e^x \cdot x + C \)
B \( e^{x+1} + C \)
C \( \dfrac{e^x}{x} + C \)
D \( e^x + C \)
✅ Solution The exponential function \(e^x\) is its own derivative and hence its own anti-derivative. \(\displaystyle\int e^x\,dx = e^x + C\).
Answer: (D) \( e^x + C \)
5
MCQ · Numerical
Using substitution, evaluate: \( \displaystyle\int 2x\sin(x^2+1)\,dx \)
A \( \sin(x^2+1) + C \)
B \( \cos(x^2+1) + C \)
C \( -\cos(x^2+1) + C \)
D \( -\sin(x^2+1) + C \)
✅ Solution Put \(t = x^2 + 1 \Rightarrow dt = 2x\,dx\).
\(\displaystyle\int 2x\sin(x^2+1)\,dx = \int \sin t\,dt = -\cos t + C = -\cos(x^2+1) + C\).
Answer: (C) \( -\cos(x^2+1)+C \)
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6
MCQ · Numerical
The value of \( \displaystyle\int_0^1 x^2\,dx \) is:
A \( 0 \)
B \( \dfrac{1}{3} \)
C \( \dfrac{1}{2} \)
D \( 1 \)
✅ Solution \(\displaystyle\int_0^1 x^2\,dx = \left[\frac{x^3}{3}\right]_0^1 = \frac{1}{3} - 0 = \frac{1}{3}\).
Answer: (B) \( \dfrac{1}{3} \)
7
MCQ · Numerical
The value of \( \displaystyle\int \tan x\,dx \) is:
A \( \log|\sin x| + C \)
B \( -\log|\cos x| + C \)
C \( \log|\sec x| + C \)
D Both (B) and (C)
✅ Solution \(\displaystyle\int \tan x\,dx = \int \frac{\sin x}{\cos x}\,dx\). Put \(\cos x = t \Rightarrow -\sin x\,dx = dt\).
\(= -\log|\cos x| + C = \log|\sec x| + C\). Both (B) and (C) are correct since \(-\log|\cos x| = \log|\sec x|\).
Answer: (D) Both (B) and (C)
8
MCQ · Numerical
Using integration by parts, \( \displaystyle\int x e^x\,dx \) equals:
A \( x e^x + e^x + C \)
B \( x e^x - e^x + C \)
C \( e^x(x-1) + C \)
D Both (B) and (C)
✅ Solution Let first function = \(x\), second = \(e^x\).
\(\displaystyle\int x e^x\,dx = x e^x - \int 1\cdot e^x\,dx = xe^x - e^x + C = e^x(x-1)+C\).
Answer: (D) Both (B) and (C)
9
MCQ · Numerical
The value of \( \displaystyle\int_0^{\pi/2} \sin x\,dx \) is:
A \( 0 \)
B \( 2 \)
C \( \pi \)
D \( 1 \)
✅ Solution \(\displaystyle\int_0^{\pi/2}\sin x\,dx = [-\cos x]_0^{\pi/2} = -\cos\tfrac{\pi}{2} + \cos 0 = 0 + 1 = 1\).
Answer: (D) \( 1 \)
10
MCQ · Numerical
The value of \( \displaystyle\int \frac{1}{x}\,dx \) is:
A \( x + C \)
B \( \log x + C \)
C \( \log |x| + C \)
D \( \dfrac{1}{x^2} + C \)
✅ Solution \(\displaystyle\int \frac{1}{x}\,dx = \log|x| + C\), valid for all \(x \neq 0\) (including negative \(x\)).
Answer: (C) \( \log|x| + C \)
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11
MCQ · Numerical
\( \displaystyle\int_0^{\pi} \sin^5 x \cos^4 x\,dx \) equals:
A \( \dfrac{8}{315} \)
B \( \dfrac{16}{315} \)
C \( 0 \)
D \( \dfrac{4}{315} \)
✅ Solution Use the reduction formula. Since \(\sin^5 x\) is odd in \([0,\pi]\) with \(\cos^4 x\) even, we use Wallis' formula: \(\displaystyle\int_0^\pi \sin^5 x\cos^4 x\,dx = 2\int_0^{\pi/2}\sin^5 x\cos^4 x\,dx\).
By the Beta function: \(= 2 \cdot \dfrac{4!\cdot 3!}{2\cdot 9!} \cdot \frac{1}{2} = \dfrac{8}{315}\).
Answer: (A) \( \dfrac{8}{315} \)
12
MCQ · Numerical
By property P₄, \( \displaystyle\int_0^a f(x)\,dx \) equals:
A \( \displaystyle\int_0^a f(a-x)\,dx \)
B \( \displaystyle\int_0^a f(x+a)\,dx \)
C \( \displaystyle-\int_0^a f(x)\,dx \)
D \( \displaystyle\int_a^0 f(x)\,dx \)
✅ Solution Property P₄ states: \(\displaystyle\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx\). This is proved by substituting \(t = a - x\).
Answer: (A) \( \displaystyle\int_0^a f(a-x)\,dx \)
13
MCQ · Numerical
The value of \( \displaystyle\int_0^{\pi/2} \log\sin x\,dx \) is:
A \( \dfrac{\pi}{2}\log 2 \)
B \( -\dfrac{\pi}{2}\log 2 \)
C \( 0 \)
D \( \pi\log 2 \)
✅ Solution Using the standard result (proven by adding \(I\) using P₄ and substitution): \(\displaystyle\int_0^{\pi/2}\log\sin x\,dx = -\frac{\pi}{2}\log 2\).
Answer: (B) \( -\dfrac{\pi}{2}\log 2 \)
14
MCQ · Numerical
Evaluate: \( \displaystyle\int \frac{dx}{x^2 - a^2} \)
A \( \dfrac{1}{2a}\log\left|\dfrac{x+a}{x-a}\right| + C \)
B \( \dfrac{1}{2a}\log\left|\dfrac{x-a}{x+a}\right| + C \)
C \( \dfrac{1}{a}\tan^{-1}\dfrac{x}{a} + C \)
D \( \log|x^2-a^2| + C \)
✅ Solution Using partial fractions: \(\dfrac{1}{x^2-a^2} = \dfrac{1}{2a}\left(\dfrac{1}{x-a}-\dfrac{1}{x+a}\right)\). Integrating: \(\dfrac{1}{2a}\log\left|\dfrac{x-a}{x+a}\right| + C\).
Answer: (B) \( \dfrac{1}{2a}\log\left|\dfrac{x-a}{x+a}\right| + C \)
15
MCQ · Numerical
If \( f(x) \) is an odd function, then \( \displaystyle\int_{-a}^{a} f(x)\,dx = \)
A \( 2\displaystyle\int_0^a f(x)\,dx \)
B \( \displaystyle\int_0^a f(x)\,dx \)
C \( 0 \)
D \( -1 \)
✅ Solution Property P₇(ii): If \(f(-x) = -f(x)\) (odd function), then \(\displaystyle\int_{-a}^a f(x)\,dx = 0\).
Answer: (C) \( 0 \)
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Section B — One-Value / One-Word Answer

Fill in the correct numerical answer for each question. (Q16–Q25)

16
One Value · Numerical
Evaluate: \( \displaystyle\int_2^3 x^2\,dx \). Write the numerical answer.
✅ Solution \(\displaystyle\int_2^3 x^2\,dx = \left[\frac{x^3}{3}\right]_2^3 = \frac{27}{3} - \frac{8}{3} = \frac{19}{3}\).
Answer: \( \dfrac{19}{3} \)
17
One Value · Numerical
The value of \( \displaystyle\int_0^{1} \frac{dx}{1+x^2} \) is \(\_\_\_\_\).
✅ Solution \(\displaystyle\int_0^1 \frac{dx}{1+x^2} = [\tan^{-1}x]_0^1 = \tan^{-1}1 - \tan^{-1}0 = \frac{\pi}{4} - 0 = \frac{\pi}{4}\).
Answer: \( \dfrac{\pi}{4} \)
18
One Value · Numerical
Evaluate: \( \displaystyle\int_0^{\pi/4} \tan x\,dx \). (Give exact form)
✅ Solution \(\displaystyle\int_0^{\pi/4}\tan x\,dx = [\log|\sec x|]_0^{\pi/4} = \log|\sec\tfrac{\pi}{4}| - \log|\sec 0|\) \(= \log\sqrt{2} - \log 1 = \log\sqrt{2} = \dfrac{1}{2}\log 2\).
Answer: \( \dfrac{1}{2}\log 2 \)
19
One Value · Numerical
Find: \( \displaystyle\int \cos^2 x\,dx \). Express in simplest form.
✅ Solution Use identity: \(\cos^2 x = \dfrac{1+\cos 2x}{2}\). \(\displaystyle\int\cos^2 x\,dx = \int\frac{1+\cos 2x}{2}\,dx = \frac{x}{2} + \frac{\sin 2x}{4} + C\).
Answer: \( \dfrac{x}{2} + \dfrac{\sin 2x}{4} + C \)
20
One Value · Numerical
The value of \( \displaystyle\int_0^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx \) is \(\_\_\_\_\).
✅ Solution Let \(I = \displaystyle\int_0^{\pi/2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx\). By P₄: \(I = \displaystyle\int_0^{\pi/2}\frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}\,dx\). Adding: \(2I = \displaystyle\int_0^{\pi/2}1\,dx = \frac{\pi}{2}\), so \(I = \frac{\pi}{4}\).
Answer: \( \dfrac{\pi}{4} \)
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21
One Value · Numerical
Evaluate: \( \displaystyle\int \frac{dx}{\sqrt{a^2-x^2}} \). Express with inverse trig.
✅ Solution Standard result (formula 5 from textbook): \(\displaystyle\int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\frac{x}{a} + C\).
Answer: \( \sin^{-1}\dfrac{x}{a} + C \)
22
One Value · Numerical
\( \displaystyle\int_1^3 \frac{1}{x}\,dx \) = \(\_\_\_\_\) (give exact log form)
✅ Solution \(\displaystyle\int_1^3 \frac{1}{x}\,dx = [\log|x|]_1^3 = \log 3 - \log 1 = \log 3 - 0 = \log 3\).
Answer: \( \log 3 \)
23
One Value · Numerical
Find the anti-derivative of \( 4x^3 - 6 \) such that \( F(0) = 3 \). The value of \( F(1) \) is \(\_\_\_\_\).
✅ Solution \(\displaystyle\int (4x^3-6)\,dx = x^4 - 6x + C\). Given \(F(0)=3 \Rightarrow 0 - 0 + C = 3 \Rightarrow C=3\). So \(F(x) = x^4 - 6x + 3\). \(F(1) = 1 - 6 + 3 = -2\).
Answer: \( -2 \)
24
One Value · Numerical
Evaluate: \( \displaystyle\int_0^{\pi/2}\sin^3 x\,dx \). Write the exact fraction.
✅ Solution Write \(\sin^3 x = (1-\cos^2 x)\sin x\). Put \(t=\cos x,\, dt=-\sin x\,dx\): \(\displaystyle\int_0^{\pi/2}\sin^3 x\,dx = \int_1^0 -(1-t^2)\,dt = \int_0^1(1-t^2)\,dt = \left[t-\frac{t^3}{3}\right]_0^1 = 1-\frac{1}{3} = \frac{2}{3}\).
Answer: \( \dfrac{2}{3} \)
25
One Value · Numerical
Using the Second Fundamental Theorem, evaluate: \( \displaystyle\int_4^9 \frac{\sqrt{x}}{(30-x^{3/2})^2}\,dx \). (Use the result from NCERT Example 25)
✅ Solution Put \(t = 30 - x^{3/2} \Rightarrow dt = -\frac{3}{2}\sqrt{x}\,dx\), so \(\sqrt{x}\,dx = -\frac{2}{3}\,dt\).
Anti-derivative: \(\dfrac{2}{3(30-x^{3/2})}\).
\(\displaystyle I = \left[\frac{2}{3(30-x^{3/2})}\right]_4^9 = \frac{2}{3}\left(\frac{1}{30-27}-\frac{1}{30-8}\right) = \frac{2}{3}\left(\frac{1}{3}-\frac{1}{22}\right) = \frac{2}{3}\cdot\frac{19}{66} = \frac{19}{99}\).
Answer: \( \dfrac{19}{99} \)
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