Exercise 7.8, Question 1
\(\displaystyle\int_{-1}^{1} (x + 1)\,dx\)
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- \[\left[\dfrac{x^2}{2} + x\right]_{-1}^{1} = \tfrac32 - \left(-\tfrac12\right)\]
Answer: \(2\)
NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.8
Exercise 7.8: Definite integrals by the fundamental theorem. Second fundamental theorem: if \(F' = f\) on \([a, b]\), then \(\int_a^b f(x)\,dx = F(b) - F(a)\); no C is needed. Find an antiderivative by any method, then substitute the limits carefully (brackets!).
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Exercise 7.8 questions and solutions
Exercise 7.8, Question 2
\(\displaystyle\int_{2}^{3} \dfrac1x\,dx\)
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- \([\log x]_2^3 = \log 3 - \log 2\).
Answer: \(\log\tfrac32\)
Exercise 7.8, Question 3
\(\displaystyle\int_{1}^{2} (4x^3 - 5x^2 + 6x + 9)\,dx\)
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- \(F = x^4 - \tfrac53x^3 + 3x^2 + 9x\).
- \[\begin{aligned}F(2) &= 16 - \tfrac{40}{3} + 12 + 18 \\ &= \tfrac{98}{3}\end{aligned}\]; \[\begin{aligned}F(1) &= 1 - \tfrac53 + 3 + 9 \\ &= \tfrac{34}{3}\end{aligned}\]
Answer: \(\tfrac{64}{3}\)
Exercise 7.8, Question 4
\(\displaystyle\int_{0}^{\frac{\pi}{4}} \sin 2x\,dx\)
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- \[\left[-\dfrac{\cos 2x}{2}\right]_0^{\pi/4} = 0 + \tfrac12\]
Answer: \(\tfrac12\)
Exercise 7.8, Question 5
\(\displaystyle\int_{0}^{\frac{\pi}{2}} \cos 2x\,dx\)
Show solution
- \[\left[\dfrac{\sin 2x}{2}\right]_0^{\pi/2} = \dfrac{\sin\pi - \sin 0}{2}\]
Answer: \(0\)
Exercise 7.8, Question 6
\(\displaystyle\int_{4}^{5} e^x\,dx\)
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- \(e^5 - e^4\).
Answer: \(e^4(e - 1)\)
Exercise 7.8, Question 7
\(\displaystyle\int_{0}^{\frac{\pi}{4}} \tan x\,dx\)
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- \[[\log|\sec x|]_0^{\pi/4} = \log\sqrt2 - \log 1\]
Answer: \(\tfrac12\log 2\)
Exercise 7.8, Question 8
\(\displaystyle\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \csc x\,dx\)
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- \(\int\csc x\,dx = \log|\csc x - \cot x|\).
- At \(\tfrac{\pi}{4}\): \(\sqrt2 - 1\). At \(\tfrac{\pi}{6}\): \(2 - \sqrt3\).
Answer: \(\log\dfrac{\sqrt2 - 1}{2 - \sqrt3}\)
Exercise 7.8, Question 9
\(\displaystyle\int_{0}^{1} \dfrac{1}{\sqrt{1 - x^2}}\,dx\)
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- \([\sin^{-1}x]_0^1 = \tfrac{\pi}{2} - 0\).
Answer: \(\tfrac{\pi}{2}\)
Exercise 7.8, Question 10
\(\displaystyle\int_{0}^{1} \dfrac{1}{1 + x^2}\,dx\)
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- \([\tan^{-1}x]_0^1\).
Answer: \(\tfrac{\pi}{4}\)
Exercise 7.8, Question 11
\(\displaystyle\int_{2}^{3} \dfrac{1}{x^2 - 1}\,dx\)
Show solution
- \[\left[\tfrac12\log\left|\dfrac{x - 1}{x + 1}\right|\right]_2^3 = \tfrac12\left(\log\tfrac12 - \log\tfrac13\right)\]
Answer: \(\tfrac12\log\tfrac32\)
Exercise 7.8, Question 12
\(\displaystyle\int_{0}^{\frac{\pi}{2}} \cos^2 x\,dx\)
Show solution
- \(\cos^2 x = \tfrac12(1 + \cos 2x)\): \[\left[\dfrac{x}{2} + \dfrac{\sin 2x}{4}\right]_0^{\pi/2}\]
Answer: \(\tfrac{\pi}{4}\)
Exercise 7.8, Question 13
\(\displaystyle\int_{2}^{3} \dfrac{x}{x^2 + 1}\,dx\)
Show solution
- \[\left[\tfrac12\log(x^2 + 1)\right]_2^3 = \tfrac12\log\tfrac{10}{5}\]
Answer: \(\tfrac12\log 2\)
Exercise 7.8, Question 14
\(\displaystyle\int_{0}^{1} \dfrac{2x + 3}{5x^2 + 1}\,dx\)
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- \[\int\dfrac{2x}{5x^2 + 1}\,dx = \tfrac15\log(5x^2 + 1)\]; \[\int\dfrac{3}{5x^2 + 1}\,dx = \dfrac{3}{\sqrt5}\tan^{-1}(\sqrt5\,x)\]
Answer: \[\tfrac15\log 6 + \dfrac{3}{\sqrt5}\tan^{-1}\sqrt5\]
Exercise 7.8, Question 15
\(\displaystyle\int_{0}^{1} xe^{x^2}\,dx\)
Show solution
- \(\left[\tfrac12e^{x^2}\right]_0^1\).
Answer: \(\tfrac12(e - 1)\)
Exercise 7.8, Question 16
\(\displaystyle\int_{1}^{2} \dfrac{5x^2}{x^2 + 4x + 3}\,dx\)
Show solution
- Divide: \(5 - \dfrac{20x + 15}{(x + 1)(x + 3)}\), and \[\dfrac{20x + 15}{(x + 1)(x + 3)} = \dfrac{-5/2}{x + 1} + \dfrac{45/2}{x + 3}\]
- \[\left[5x + \tfrac52\log(x + 1) - \tfrac{45}{2}\log(x + 3)\right]_1^2 = 5 + \tfrac52\log\tfrac32 - \tfrac{45}{2}\log\tfrac54\]
Answer: \[5 - \tfrac52\left[9\log\tfrac54 - \log\tfrac32\right]\]
Exercise 7.8, Question 17
\(\displaystyle\int_{0}^{\frac{\pi}{4}} (2\sec^2 x + x^3 + 2)\,dx\)
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- \[\left[2\tan x + \dfrac{x^4}{4} + 2x\right]_0^{\pi/4} = 2 + \dfrac{\pi^4}{4 \cdot 256} + \dfrac{\pi}{2}\]
Answer: \(2 + \dfrac{\pi}{2} + \dfrac{\pi^4}{1024}\)
Exercise 7.8, Question 18
\(\displaystyle\int_{0}^{\pi} \left(\sin^2\dfrac{x}{2} - \cos^2\dfrac{x}{2}\right)\,dx\)
Show solution
- The integrand is \(-\cos x\): \([-\sin x]_0^{\pi} = 0\).
Answer: \(0\)
Exercise 7.8, Question 19
\(\displaystyle\int_{0}^{2} \dfrac{6x + 3}{x^2 + 4}\,dx\)
Show solution
- \[\int\dfrac{6x}{x^2 + 4}\,dx = 3\log(x^2 + 4)\]; \[\int\dfrac{3}{x^2 + 4}\,dx = \tfrac32\tan^{-1}\dfrac{x}{2}\]
- \[3\log\tfrac84 + \tfrac32 \cdot \tfrac{\pi}{4}\]
Answer: \(3\log 2 + \dfrac{3\pi}{8}\)
Exercise 7.8, Question 20
\(\displaystyle\int_{0}^{1} \left(xe^x + \sin\dfrac{\pi x}{4}\right)\,dx\)
Show solution
- \(\int xe^x\,dx = e^x(x - 1)\): from 0 to 1 gives \(0 - (-1) = 1\).
- \[\left[-\dfrac{4}{\pi}\cos\dfrac{\pi x}{4}\right]_0^1 = \dfrac{4}{\pi}\left(1 - \dfrac{1}{\sqrt2}\right)\]
Answer: \[1 + \dfrac{4}{\pi} - \dfrac{2\sqrt2}{\pi}\]
Exercise 7.8, Question 21
\(\displaystyle\int_1^{\sqrt3}\dfrac{dx}{1 + x^2}\) equals: (A) \(\tfrac{\pi}{3}\) (B) \(\tfrac{2\pi}{3}\) (C) \(\tfrac{\pi}{6}\) (D) \(\tfrac{\pi}{12}\)
Show solution
- \[\tan^{-1}\sqrt3 - \tan^{-1}1 = \tfrac{\pi}{3} - \tfrac{\pi}{4}\]
Answer: (D)
Exercise 7.8, Question 22
\(\displaystyle\int_0^{2/3}\dfrac{dx}{4 + 9x^2}\) equals: (A) \(\tfrac{\pi}{6}\) (B) \(\tfrac{\pi}{12}\) (C) \(\tfrac{\pi}{24}\) (D) \(\tfrac{\pi}{4}\)
Show solution
- \[\int\dfrac{dx}{4 + 9x^2} = \tfrac16\tan^{-1}\dfrac{3x}{2}\]; at \(\tfrac23\): \(\tfrac16 \cdot \tfrac{\pi}{4}\).
Answer: (C)
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