Exercise 7.1, Question 1
\(\displaystyle\int \sin 2x\,dx\)
Show solution
- \(\dfrac{d}{dx}(\cos 2x) = -2\sin 2x\), so divide by \(-2\).
Answer: \(-\tfrac12\cos 2x + C\)
NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.1
Exercise 7.1: Integration as the inverse of differentiation. Read each standard derivative backwards: \(\int x^n\,dx = \dfrac{x^{n+1}}{n+1}\) (\(n \ne -1\)), \(\int\dfrac{dx}{x} = \log|x|\), \(\int e^x\,dx = e^x\), \(\int\cos x\,dx = \sin x\), \(\int\sec^2 x\,dx = \tan x\), \(\int\sec x\tan x\,dx = \sec x\); for \(f(ax + b)\) divide by a. Split sums, expand products and divide out fractions first. Always add C.
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Exercise 7.1 questions and solutions
Exercise 7.1, Question 2
\(\displaystyle\int \cos 3x\,dx\)
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- \(\dfrac{d}{dx}(\sin 3x) = 3\cos 3x\).
Answer: \(\tfrac13\sin 3x + C\)
Exercise 7.1, Question 3
\(\displaystyle\int e^{2x}\,dx\)
Show solution
- \(\dfrac{d}{dx}e^{2x} = 2e^{2x}\).
Answer: \(\tfrac12e^{2x} + C\)
Exercise 7.1, Question 4
\(\displaystyle\int (ax + b)^2\,dx\)
Show solution
- \(\dfrac{d}{dx}(ax + b)^3 = 3a(ax + b)^2\).
Answer: \(\dfrac{(ax + b)^3}{3a} + C\)
Exercise 7.1, Question 5
\(\displaystyle\int (\sin 2x - 4e^{3x})\,dx\)
Show solution
- Term by term: \(\int\sin 2x\,dx = -\tfrac12\cos 2x\), \(\int e^{3x}\,dx = \tfrac13e^{3x}\).
Answer: \(-\tfrac12\cos 2x - \tfrac43e^{3x} + C\)
Exercise 7.1, Question 6
\(\displaystyle\int (4e^{3x} + 1)\,dx\)
Show solution
- \(4 \cdot \tfrac13e^{3x} + x\).
Answer: \(\tfrac43e^{3x} + x + C\)
Exercise 7.1, Question 7
\(\displaystyle\int x^2\left(1 - \dfrac{1}{x^2}\right)\,dx\)
Show solution
- Expand: \(x^2 - 1\).
Answer: \(\dfrac{x^3}{3} - x + C\)
Exercise 7.1, Question 8
\(\displaystyle\int (ax^2 + bx + c)\,dx\)
Show solution
- Power rule term by term.
Answer: \[\dfrac{ax^3}{3} + \dfrac{bx^2}{2} + cx + C\]
Exercise 7.1, Question 9
\(\displaystyle\int (2x^2 + e^x)\,dx\)
Show solution
- Power rule and \(\int e^x\,dx = e^x\).
Answer: \(\dfrac{2x^3}{3} + e^x + C\)
Exercise 7.1, Question 10
\(\displaystyle\int \left(\sqrt x - \dfrac{1}{\sqrt x}\right)^2\,dx\)
Show solution
- Expand: \(x - 2 + \dfrac1x\).
Answer: \(\dfrac{x^2}{2} - 2x + \log|x| + C\)
Exercise 7.1, Question 11
\(\displaystyle\int \dfrac{x^3 + 5x^2 - 4}{x^2}\,dx\)
Show solution
- Divide: \(x + 5 - 4x^{-2}\); \(\int -4x^{-2}\,dx = 4x^{-1}\).
Answer: \(\dfrac{x^2}{2} + 5x + \dfrac4x + C\)
Exercise 7.1, Question 12
\(\displaystyle\int \dfrac{x^3 + 3x + 4}{\sqrt x}\,dx\)
Show solution
- Divide: \(x^{5/2} + 3x^{1/2} + 4x^{-1/2}\).
Answer: \[\tfrac27x^{7/2} + 2x^{3/2} + 8\sqrt x + C\]
Exercise 7.1, Question 13
\(\displaystyle\int \dfrac{x^3 - x^2 + x - 1}{x - 1}\,dx\)
Show solution
- \[\begin{aligned}x^3 - x^2 + x - 1 &= x^2(x - 1) + (x - 1) \\ &= (x - 1)(x^2 + 1)\end{aligned}\], so the integrand is \(x^2 + 1\).
Answer: \(\dfrac{x^3}{3} + x + C\)
Exercise 7.1, Question 14
\(\displaystyle\int (1 - x)\sqrt x\,dx\)
Show solution
- \(x^{1/2} - x^{3/2}\).
Answer: \(\tfrac23x^{3/2} - \tfrac25x^{5/2} + C\)
Exercise 7.1, Question 15
\(\displaystyle\int \sqrt x(3x^2 + 2x + 3)\,dx\)
Show solution
- \(3x^{5/2} + 2x^{3/2} + 3x^{1/2}\).
Answer: \[\tfrac67x^{7/2} + \tfrac45x^{5/2} + 2x^{3/2} + C\]
Exercise 7.1, Question 16
\(\displaystyle\int (2x - 3\cos x + e^x)\,dx\)
Show solution
- Term by term.
Answer: \(x^2 - 3\sin x + e^x + C\)
Exercise 7.1, Question 17
\(\displaystyle\int (2x^2 - 3\sin x + 5\sqrt x)\,dx\)
Show solution
- \(\int -3\sin x\,dx = 3\cos x\); \(\int 5x^{1/2}\,dx = \tfrac{10}{3}x^{3/2}\).
Answer: \[\tfrac23x^3 + 3\cos x + \tfrac{10}{3}x^{3/2} + C\]
Exercise 7.1, Question 18
\(\displaystyle\int \sec x(\sec x + \tan x)\,dx\)
Show solution
- \(\sec^2 x + \sec x\tan x\).
Answer: \(\tan x + \sec x + C\)
Exercise 7.1, Question 19
\(\displaystyle\int \dfrac{\sec^2 x}{\csc^2 x}\,dx\)
Show solution
- \[\begin{aligned}\dfrac{\sec^2 x}{\csc^2 x} &= \tan^2 x \\ &= \sec^2 x - 1\end{aligned}\]
Answer: \(\tan x - x + C\)
Exercise 7.1, Question 20
\(\displaystyle\int \dfrac{2 - 3\sin x}{\cos^2 x}\,dx\)
Show solution
- \(2\sec^2 x - 3\sec x\tan x\).
Answer: \(2\tan x - 3\sec x + C\)
Exercise 7.1, Question 21
The antiderivative of \(\sqrt x + \dfrac{1}{\sqrt x}\) is: (A) \(\tfrac13x^{1/3} + 2x^{1/2} + C\) (B) \(\tfrac23x^{2/3} + \tfrac12x^2 + C\) (C) \(\tfrac23x^{3/2} + 2x^{1/2} + C\) (D) \(\tfrac32x^{3/2} + \tfrac12x^{1/2} + C\)
Show solution
- \(\int x^{1/2}\,dx = \tfrac23x^{3/2}\), \(\int x^{-1/2}\,dx = 2x^{1/2}\).
Answer: (C)
Exercise 7.1, Question 22
\(\dfrac{d}{dx}f(x) = 4x^3 - \dfrac{3}{x^4}\) and \(f(2) = 0\). Then \(f(x)\) is: (A) \(x^4 + \dfrac{1}{x^3} - \dfrac{129}{8}\) (B) \(x^3 + \dfrac{1}{x^4} + \dfrac{129}{8}\) (C) \(x^4 + \dfrac{1}{x^3} + \dfrac{129}{8}\) (D) \(x^3 + \dfrac{1}{x^4} - \dfrac{129}{8}\)
Show solution
- \(f(x) = x^4 + x^{-3} + C\).
- \[\begin{aligned}&f(2) = 16 + \tfrac18 + C = 0 \\ \Rightarrow\ &C = -\tfrac{129}{8}\end{aligned}\]
Answer: (A)
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