NCERT Solutions Class-12-Chapter-7-Integrals
Excercise-7.8
Note: The Second Fundamental Theorem of Calculus states that if \( F \) is an antiderivative of \( f \), then \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \).
Q1
Evaluate the definite integral: \( \int_{-1}^{1} (x+1) \, dx \)
Solution:
First, find the antiderivative of \( x+1 \): \( \int (x+1) \, dx = \frac{x^2}{2} + x = F(x) \).
Apply the limits:
\[ \begin{aligned} \int_{-1}^{1} (x+1) \, dx &= F(1) - F(-1) \\ &= \left[ \frac{1^2}{2} + 1 \right] - \left[ \frac{(-1)^2}{2} + (-1) \right] \\ &= \left( \frac{1}{2} + 1 \right) - \left( \frac{1}{2} - 1 \right) \\ &= \frac{3}{2} - \left( -\frac{1}{2} \right) \\ &= \frac{3}{2} + \frac{1}{2} = 2 \end{aligned} \]
Q2
Evaluate the definite integral: \( \int_{2}^{3} \frac{1}{x} \, dx \)
Solution:
Antiderivative is \( \log |x| \).
\[ \begin{aligned} \int_{2}^{3} \frac{1}{x} \, dx &= [\log x]_{2}^{3} \\ &= \log 3 - \log 2 \\ &= \log \left( \frac{3}{2} \right) \end{aligned} \]
Q3
Evaluate the definite integral: \( \int_{1}^{2} (4x^3 - 5x^2 + 6x + 9) \, dx \)
Solution:
\[ \begin{aligned} I &= \left[ 4\frac{x^4}{4} - 5\frac{x^3}{3} + 6\frac{x^2}{2} + 9x \right]_{1}^{2} \\ &= \left[ x^4 - \frac{5}{3}x^3 + 3x^2 + 9x \right]_{1}^{2} \\ &= \left( 2^4 - \frac{5}{3}(8) + 3(4) + 18 \right) - \left( 1 - \frac{5}{3} + 3 + 9 \right) \\ &= \left( 16 - \frac{40}{3} + 12 + 18 \right) - \left( 13 - \frac{5}{3} \right) \\ &= \left( 46 - \frac{40}{3} \right) - \left( \frac{34}{3} \right) \\ &= 46 - \frac{74}{3} \\ &= \frac{138 - 74}{3} = \frac{64}{3} \end{aligned} \]
Q4
Evaluate the definite integral: \( \int_{0}^{\frac{\pi}{4}} \sin 2x \, dx \)
Solution:
\[ \begin{aligned} I &= \left[ \frac{-\cos 2x}{2} \right]_{0}^{\frac{\pi}{4}} \\ &= -\frac{1}{2} \left[ \cos \left( 2 \cdot \frac{\pi}{4} \right) - \cos 0 \right] \\ &= -\frac{1}{2} \left[ \cos \frac{\pi}{2} - 1 \right] \\ &= -\frac{1}{2} (0 - 1) = \frac{1}{2} \end{aligned} \]
Q5
Evaluate the definite integral: \( \int_{0}^{\frac{\pi}{2}} \cos 2x \, dx \)
Solution:
\[ \begin{aligned} I &= \left[ \frac{\sin 2x}{2} \right]_{0}^{\frac{\pi}{2}} \\ &= \frac{1}{2} \left[ \sin \left( 2 \cdot \frac{\pi}{2} \right) - \sin 0 \right] \\ &= \frac{1}{2} (\sin \pi - 0) \\ &= \frac{1}{2} (0 - 0) = 0 \end{aligned} \]
Q6
Evaluate the definite integral: \( \int_{4}^{5} e^x \, dx \)
Solution:
\[ \begin{aligned} I &= [e^x]_{4}^{5} \\ &= e^5 - e^4 \\ &= e^4(e - 1) \end{aligned} \]
Q7
Evaluate the definite integral: \( \int_{0}^{\frac{\pi}{4}} \tan x \, dx \)
Solution:
\[ \begin{aligned} I &= [\log |\sec x|]_{0}^{\frac{\pi}{4}} \\ &= \log |\sec \frac{\pi}{4}| - \log |\sec 0| \\ &= \log \sqrt{2} - \log 1 \\ &= \log (2^{1/2}) - 0 \\ &= \frac{1}{2} \log 2 \end{aligned} \]
Q8
Evaluate the definite integral: \( \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \text{cosec } x \, dx \)
Solution:
\[ \begin{aligned} I &= [\log |\text{cosec } x - \cot x|]_{\frac{\pi}{6}}^{\frac{\pi}{4}} \\ &= \log |\text{cosec } \frac{\pi}{4} - \cot \frac{\pi}{4}| - \log |\text{cosec } \frac{\pi}{6} - \cot \frac{\pi}{6}| \\ &= \log |\sqrt{2} - 1| - \log |2 - \sqrt{3}| \\ &= \log \left( \frac{\sqrt{2}-1}{2-\sqrt{3}} \right) \end{aligned} \]
Q9
Evaluate the definite integral: \( \int_{0}^{1} \frac{dx}{\sqrt{1-x^2}} \)
Solution:
\[ \begin{aligned} I &= [\sin^{-1} x]_{0}^{1} \\ &= \sin^{-1} 1 - \sin^{-1} 0 \\ &= \frac{\pi}{2} - 0 = \frac{\pi}{2} \end{aligned} \]
Q10
Evaluate the definite integral: \( \int_{0}^{1} \frac{dx}{1+x^2} \)
Solution:
\[ \begin{aligned} I &= [\tan^{-1} x]_{0}^{1} \\ &= \tan^{-1} 1 - \tan^{-1} 0 \\ &= \frac{\pi}{4} - 0 = \frac{\pi}{4} \end{aligned} \]
Q11
Evaluate the definite integral: \( \int_{2}^{3} \frac{dx}{x^2-1} \)
Solution:
Using formula \( \int \frac{dx}{x^2-a^2} = \frac{1}{2a} \log \left| \frac{x-a}{x+a} \right| \):
\[ \begin{aligned} I &= \frac{1}{2} \left[ \log \left| \frac{x-1}{x+1} \right| \right]_{2}^{3} \\ &= \frac{1}{2} \left( \log \frac{2}{4} - \log \frac{1}{3} \right) \\ &= \frac{1}{2} \left( \log \frac{1}{2} - \log \frac{1}{3} \right) \\ &= \frac{1}{2} \log \left( \frac{1/2}{1/3} \right) \\ &= \frac{1}{2} \log \frac{3}{2} \end{aligned} \]
Q12
Evaluate the definite integral: \( \int_{0}^{\frac{\pi}{2}} \cos^2 x \, dx \)
Solution:
Use \( \cos^2 x = \frac{1+\cos 2x}{2} \).
\[ \begin{aligned} I &= \frac{1}{2} \int_{0}^{\frac{\pi}{2}} (1 + \cos 2x) \, dx \\ &= \frac{1}{2} \left[ x + \frac{\sin 2x}{2} \right]_{0}^{\frac{\pi}{2}} \\ &= \frac{1}{2} \left[ \left( \frac{\pi}{2} + \frac{\sin \pi}{2} \right) - (0 + 0) \right] \\ &= \frac{1}{2} \left( \frac{\pi}{2} + 0 \right) = \frac{\pi}{4} \end{aligned} \]
Q13
Evaluate the definite integral: \( \int_{2}^{3} \frac{x}{x^2+1} \, dx \)
Solution:
Let \( x^2+1 = t \). Then \( 2x dx = dt \Rightarrow x dx = dt/2 \).
Limits: \( x=2 \to t=5 \); \( x=3 \to t=10 \).
\[ \begin{aligned} I &= \frac{1}{2} \int_{5}^{10} \frac{dt}{t} \\ &= \frac{1}{2} [\log t]_{5}^{10} \\ &= \frac{1}{2} (\log 10 - \log 5) \\ &= \frac{1}{2} \log \frac{10}{5} = \frac{1}{2} \log 2 = \log \sqrt{2} \end{aligned} \]
Q14
Evaluate the definite integral: \( \int_{0}^{1} \frac{2x+3}{5x^2+1} \, dx \)
Solution:
Split the integral: \( \int \frac{2x}{5x^2+1} dx + 3 \int \frac{1}{5x^2+1} dx \).
Part 1: \( \frac{1}{5} \int \frac{10x}{5x^2+1} dx = \frac{1}{5} [\log(5x^2+1)]_0^1 = \frac{1}{5} \log 6 \).
Part 2: \( 3 \int \frac{dx}{(\sqrt{5}x)^2+1} = \frac{3}{\sqrt{5}} [\tan^{-1}(\sqrt{5}x)]_0^1 = \frac{3}{\sqrt{5}} \tan^{-1}\sqrt{5} \).
Result: \( \frac{1}{5} \log 6 + \frac{3}{\sqrt{5}} \tan^{-1}\sqrt{5} \).
Q15
Evaluate the definite integral: \( \int_{0}^{1} x e^{x^2} \, dx \)
Solution:
Let \( x^2 = t \Rightarrow 2x dx = dt \). Limits: \( 0 \to 0, 1 \to 1 \).
\[ \begin{aligned} I &= \frac{1}{2} \int_{0}^{1} e^t \, dt \\ &= \frac{1}{2} [e^t]_{0}^{1} \\ &= \frac{1}{2} (e^1 - e^0) = \frac{1}{2} (e-1) \end{aligned} \]
Q16
Evaluate the definite integral: \( \int_{1}^{2} \frac{5x^2}{x^2+4x+3} \, dx \)
Solution:
Rewrite integrand: \( 5 \left( 1 - \frac{4x+3}{x^2+4x+3} \right) = 5 - \frac{20x+15}{(x+1)(x+3)} \).
Partial Fractions: \( \frac{20x+15}{(x+1)(x+3)} = \frac{-5/2}{x+1} + \frac{45/2}{x+3} \).
Integral: \( \int_{1}^{2} \left( 5 + \frac{5}{2(x+1)} - \frac{45}{2(x+3)} \right) \, dx \).
Result: \( 5 + \frac{5}{2} \log 3 - \frac{45}{2} \log 5 + \frac{85}{2} \log 2 \).
Q17
Evaluate the definite integral: \( \int_{0}^{\frac{\pi}{4}} (2\sec^2 x + x^3 + 2) \, dx \)
Solution:
\[ \begin{aligned} I &= \left[ 2\tan x + \frac{x^4}{4} + 2x \right]_{0}^{\frac{\pi}{4}} \\ &= 2\tan\frac{\pi}{4} + \frac{1}{4}\left(\frac{\pi}{4}\right)^4 + 2\left(\frac{\pi}{4}\right) \\ &= 2(1) + \frac{\pi^4}{1024} + \frac{\pi}{2} \\ &= 2 + \frac{\pi}{2} + \frac{\pi^4}{1024} \end{aligned} \]
Q18
Evaluate the definite integral: \( \int_{0}^{\pi} (\sin^2 \frac{x}{2} - \cos^2 \frac{x}{2}) \, dx \)
Solution:
\( \sin^2 \frac{x}{2} - \cos^2 \frac{x}{2} = -\cos x \).
\[ \begin{aligned} I &= \int_{0}^{\pi} (-\cos x) \, dx \\ &= [-\sin x]_{0}^{\pi} \\ &= -(\sin \pi - \sin 0) = 0 \end{aligned} \]
Q19
Evaluate the definite integral: \( \int_{0}^{2} \frac{6x+3}{x^2+4} \, dx \)
Solution:
Split: \( \int_{0}^{2} \frac{6x}{x^2+4} dx + \int_{0}^{2} \frac{3}{x^2+4} dx \).
Part 1: \( 3[\log(x^2+4)]_0^2 = 3(\log 8 - \log 4) = 3 \log 2 \).
Part 2: \( \frac{3}{2} [\tan^{-1} \frac{x}{2}]_0^2 = \frac{3}{2} (\tan^{-1} 1 - 0) = \frac{3\pi}{8} \).
Result: \( 3 \log 2 + \frac{3\pi}{8} \).
Q20
Evaluate the definite integral: \( \int_{0}^{1} (xe^x + \sin \frac{\pi x}{4}) \, dx \)
Solution:
Part 1: \( \int_0^1 xe^x dx = [xe^x - e^x]_0^1 = (e-e) - (0-1) = 1 \).
Part 2: \( \int_0^1 \sin \frac{\pi x}{4} dx = [-\frac{4}{\pi} \cos \frac{\pi x}{4}]_0^1 = -\frac{4}{\pi} (\frac{1}{\sqrt{2}} - 1) = \frac{4}{\pi} - \frac{2\sqrt{2}}{\pi} \).
Total: \( 1 + \frac{4}{\pi} - \frac{2\sqrt{2}}{\pi} \).
Q21
\( \int_{1}^{\sqrt{3}} \frac{dx}{1+x^2} \) equals- \( \frac{\pi}{3} \)
- \( \frac{2\pi}{3} \)
- \( \frac{\pi}{6} \)
- \( \frac{\pi}{12} \)
- \( \frac{\pi}{3} \)
- \( \frac{2\pi}{3} \)
- \( \frac{\pi}{6} \)
- \( \frac{\pi}{12} \)
Answer: (D)
Solution:
\( [\tan^{-1} x]_1^{\sqrt{3}} = \frac{\pi}{3} - \frac{\pi}{4} = \frac{\pi}{12} \).
Q22
\( \int_{0}^{\frac{2}{3}} \frac{dx}{4+9x^2} \) equals- \( \frac{\pi}{6} \)
- \( \frac{\pi}{12} \)
- \( \frac{\pi}{24} \)
- \( \frac{\pi}{4} \)
- \( \frac{\pi}{6} \)
- \( \frac{\pi}{12} \)
- \( \frac{\pi}{24} \)
- \( \frac{\pi}{4} \)
Answer: (C)
Solution:
Rewrite: \( \frac{1}{9} \int_{0}^{2/3} \frac{dx}{x^2 + (2/3)^2} \).
\( = \frac{1}{9} \cdot \frac{3}{2} [\tan^{-1} \frac{3x}{2}]_0^{2/3} = \frac{1}{6} \tan^{-1} 1 = \frac{\pi}{24} \).