Download Ex. 7.9 PDF
- Ex. 7.1
- Ex. 7.2
- Ex. 7.2
- Ex. 7.3
- Ex. 7.4
- Ex. 7.5
- Ex. 7.6
- Ex. 7.7
- Ex. 7.8
- Ex. 7.9
- Ex. 7.10
- Ex. 7.11
- Miscellaneous Exercise
Exercise 7.9 Solutions
Topic: Definite Integrals
Evaluate the following definite integrals:
Question 1. Evaluate \(\int_{-1}^{1} (x+1) dx\)
Let \(I = \int_{-1}^{1} (x+1) dx\)
Integration of \(x^n\) is \(\frac{x^{n+1}}{n+1}\).
$$I = \left[ \frac{x^2}{2} + x \right]_{-1}^{1}$$
Applying limits (Upper Limit - Lower Limit):
$$I = \left( \frac{1^2}{2} + 1 \right) - \left( \frac{(-1)^2}{2} + (-1) \right)$$
$$I = \left( \frac{1}{2} + 1 \right) - \left( \frac{1}{2} - 1 \right)$$
$$I = \frac{3}{2} - \left( -\frac{1}{2} \right)$$
$$I = \frac{3}{2} + \frac{1}{2} = \frac{4}{2} = 2$$
Answer: 2
Question 2. Evaluate \(\int_{2}^{3} \frac{1}{x} dx\)
Let \(I = \int_{2}^{3} \frac{1}{x} dx\)
We know that \(\int \frac{1}{x} dx = \log |x|\).
$$I = \left[ \log |x| \right]_{2}^{3}$$
Applying limits:
$$I = \log 3 - \log 2$$
$$I = \log \left( \frac{3}{2} \right)$$
Answer: \(\log \frac{3}{2}\)
Question 3. Evaluate \(\int_{1}^{2} (4x^3 - 5x^2 + 6x + 9) dx\)
Let \(I = \int_{1}^{2} (4x^3 - 5x^2 + 6x + 9) dx\)
Integrating term by term:
$$I = \left[ 4\frac{x^4}{4} - 5\frac{x^3}{3} + 6\frac{x^2}{2} + 9x \right]_{1}^{2}$$
$$I = \left[ x^4 - \frac{5}{3}x^3 + 3x^2 + 9x \right]_{1}^{2}$$
Applying limits:
$$I = \left( 2^4 - \frac{5(2)^3}{3} + 3(2)^2 + 9(2) \right) - \left( 1^4 - \frac{5(1)^3}{3} + 3(1)^2 + 9(1) \right)$$
$$I = \left( 16 - \frac{40}{3} + 12 + 18 \right) - \left( 1 - \frac{5}{3} + 3 + 9 \right)$$
$$I = \left( 46 - \frac{40}{3} \right) - \left( 13 - \frac{5}{3} \right)$$
$$I = 46 - 13 - \frac{40}{3} + \frac{5}{3}$$
$$I = 33 - \frac{35}{3} = \frac{99 - 35}{3} = \frac{64}{3}$$
Answer: \(\frac{64}{3}\)
Question 4. Evaluate \(\int_{0}^{\pi/4} \sin 2x dx\)
Let \(I = \int_{0}^{\pi/4} \sin 2x dx\)
Integration of \(\sin ax\) is \(-\frac{\cos ax}{a}\).
$$I = \left[ -\frac{\cos 2x}{2} \right]_{0}^{\pi/4}$$
$$I = -\frac{1}{2} \left[ \cos 2(\frac{\pi}{4}) - \cos 2(0) \right]$$
$$I = -\frac{1}{2} \left[ \cos \frac{\pi}{2} - \cos 0 \right]$$
$$I = -\frac{1}{2} (0 - 1) = \frac{1}{2}$$
Answer: \(\frac{1}{2}\)
Question 5. Evaluate \(\int_{0}^{\pi/2} \cos 2x dx\)
Let \(I = \int_{0}^{\pi/2} \cos 2x dx\)
Integration of \(\cos ax\) is \(\frac{\sin ax}{a}\).
$$I = \left[ \frac{\sin 2x}{2} \right]_{0}^{\pi/2}$$
$$I = \frac{1}{2} \left[ \sin 2(\frac{\pi}{2}) - \sin 2(0) \right]$$
$$I = \frac{1}{2} (\sin \pi - \sin 0)$$
$$I = \frac{1}{2} (0 - 0) = 0$$
Answer: 0
Question 6. Evaluate \(\int_{4}^{5} e^x dx\)
Let \(I = \int_{4}^{5} e^x dx\)
$$I = \left[ e^x \right]_{4}^{5}$$
$$I = e^5 - e^4$$
$$I = e^4(e - 1)$$
Answer: \(e^4(e - 1)\)
Question 7. Evaluate \(\int_{0}^{\pi/4} \tan x dx\)
Let \(I = \int_{0}^{\pi/4} \tan x dx\)
We know \(\int \tan x dx = -\log|\cos x|\) or \(\log|\sec x|\).
$$I = \left[ -\log|\cos x| \right]_{0}^{\pi/4}$$
$$I = - [\log(\cos \frac{\pi}{4}) - \log(\cos 0)]$$
$$I = - [\log(\frac{1}{\sqrt{2}}) - \log(1)]$$
$$I = - [\log(2^{-1/2}) - 0]$$
$$I = - [-\frac{1}{2}\log 2] = \frac{1}{2}\log 2$$
Answer: \(\frac{1}{2}\log 2\)
Question 8. Evaluate \(\int_{\pi/6}^{\pi/4} \text{cosec } x dx\)
Let \(I = \int_{\pi/6}^{\pi/4} \text{cosec } x dx\)
Formula: \(\int \text{cosec } x dx = \log|\text{cosec } x - \cot x|\)
$$I = \left[ \log|\text{cosec } x - \cot x| \right]_{\pi/6}^{\pi/4}$$
$$I = \log|\text{cosec } \frac{\pi}{4} - \cot \frac{\pi}{4}| - \log|\text{cosec } \frac{\pi}{6} - \cot \frac{\pi}{6}|$$
$$I = \log|\sqrt{2} - 1| - \log|2 - \sqrt{3}|$$
$$I = \log \left( \frac{\sqrt{2} - 1}{2 - \sqrt{3}} \right)$$
Question 9. Evaluate \(\int_{0}^{1} \frac{dx}{\sqrt{1-x^2}}\)
Formula: \(\int \frac{dx}{\sqrt{1-x^2}} = \sin^{-1} x\)
$$I = \left[ \sin^{-1} x \right]_{0}^{1}$$
$$I = \sin^{-1}(1) - \sin^{-1}(0)$$
$$I = \frac{\pi}{2} - 0 = \frac{\pi}{2}$$
Answer: \(\frac{\pi}{2}\)
Question 10. Evaluate \(\int_{0}^{1} \frac{dx}{1+x^2}\)
Formula: \(\int \frac{dx}{1+x^2} = \tan^{-1} x\)
$$I = \left[ \tan^{-1} x \right]_{0}^{1}$$
$$I = \tan^{-1}(1) - \tan^{-1}(0)$$
$$I = \frac{\pi}{4} - 0 = \frac{\pi}{4}$$
Answer: \(\frac{\pi}{4}\)
Question 11. Evaluate \(\int_{2}^{3} \frac{dx}{x^2 - 1}\)
Formula: \(\int \frac{dx}{x^2 - a^2} = \frac{1}{2a} \log \left| \frac{x-a}{x+a} \right|\). Here \(a=1\).
$$I = \frac{1}{2} \left[ \log \left| \frac{x-1}{x+1} \right| \right]_{2}^{3}$$
$$I = \frac{1}{2} \left( \log \left| \frac{3-1}{3+1} \right| - \log \left| \frac{2-1}{2+1} \right| \right)$$
$$I = \frac{1}{2} \left( \log \frac{2}{4} - \log \frac{1}{3} \right)$$
$$I = \frac{1}{2} \left( \log \frac{1}{2} - \log \frac{1}{3} \right)$$
$$I = \frac{1}{2} \log \left( \frac{1/2}{1/3} \right) = \frac{1}{2} \log \left( \frac{3}{2} \right)$$
Answer: \(\frac{1}{2} \log \frac{3}{2}\)
Question 12. Evaluate \(\int_{0}^{\pi/2} \cos^2 x dx\)
Use identity: \(\cos^2 x = \frac{1 + \cos 2x}{2}\)
$$I = \int_{0}^{\pi/2} \frac{1 + \cos 2x}{2} dx$$
$$I = \frac{1}{2} \left[ x + \frac{\sin 2x}{2} \right]_{0}^{\pi/2}$$
$$I = \frac{1}{2} \left[ (\frac{\pi}{2} + \frac{\sin \pi}{2}) - (0 + \frac{\sin 0}{2}) \right]$$
$$I = \frac{1}{2} \left( \frac{\pi}{2} + 0 - 0 \right) = \frac{\pi}{4}$$
Answer: \(\frac{\pi}{4}\)
Question 13. Evaluate \(\int_{2}^{3} \frac{x dx}{x^2+1}\)
Let \(x^2 + 1 = t \Rightarrow 2x dx = dt \Rightarrow x dx = \frac{dt}{2}\)
Change limits: When \(x=2, t=5\); When \(x=3, t=10\).
$$I = \int_{5}^{10} \frac{1}{t} \frac{dt}{2} = \frac{1}{2} [\log t]_{5}^{10}$$
$$I = \frac{1}{2} (\log 10 - \log 5) = \frac{1}{2} \log(\frac{10}{5})$$
$$I = \frac{1}{2} \log 2$$
Answer: \(\frac{1}{2} \log 2\)
Question 14. Evaluate \(\int_{0}^{1} \frac{2x+3}{5x^2+1} dx\)
Split the integral: \(I = \int_{0}^{1} \frac{2x}{5x^2+1} dx + \int_{0}^{1} \frac{3}{5x^2+1} dx\)
Part 1: Multiply numerator and denominator by 5 to get derivative of denominator.
$$I_1 = \frac{1}{5} \int_{0}^{1} \frac{10x}{5x^2+1} dx = \frac{1}{5} [\log(5x^2+1)]_{0}^{1}$$
$$I_1 = \frac{1}{5} (\log 6 - \log 1) = \frac{1}{5} \log 6$$
Part 2: \(I_2 = 3 \int_{0}^{1} \frac{1}{5(x^2 + 1/5)} dx = \frac{3}{5} \int_{0}^{1} \frac{dx}{x^2 + (1/\sqrt{5})^2}\)
$$I_2 = \frac{3}{5} \cdot \frac{1}{1/\sqrt{5}} [\tan^{-1}(\frac{x}{1/\sqrt{5}})]_{0}^{1}$$
$$I_2 = \frac{3\sqrt{5}}{5} [\tan^{-1}(\sqrt{5}x)]_{0}^{1} = \frac{3}{\sqrt{5}} \tan^{-1}(\sqrt{5})$$
Total Answer: \(\frac{1}{5} \log 6 + \frac{3}{\sqrt{5}} \tan^{-1} \sqrt{5}\)
Question 15. Evaluate \(\int_{0}^{1} x e^{x^2} dx\)
Let \(x^2 = t \Rightarrow 2x dx = dt \Rightarrow x dx = \frac{dt}{2}\)
Limits: When \(x=0, t=0\); When \(x=1, t=1\).
$$I = \int_{0}^{1} e^t \frac{dt}{2} = \frac{1}{2} [e^t]_{0}^{1}$$
$$I = \frac{1}{2} (e^1 - e^0) = \frac{1}{2} (e - 1)$$
Answer: \(\frac{1}{2} (e - 1)\)
Question 16. Evaluate \(\int_{1}^{2} \frac{5x^2}{x^2+4x+3} dx\)
Divide \(5x^2\) by \(x^2+4x+3\). Quotient is 5, Remainder is \(-(20x+15)\).
$$I = \int_{1}^{2} \left( 5 - \frac{20x+15}{x^2+4x+3} \right) dx$$
$$I = [5x]_{1}^{2} - \int_{1}^{2} \frac{20x+15}{x^2+4x+3} dx$$
$$I = 5(2-1) - I_1 = 5 - I_1$$
Solving \(I_1\): Let \(20x+15 = A(2x+4) + B\). \(20x+15 = 20x + 40 - 25\).
$$I_1 = \int \frac{10(2x+4) - 25}{x^2+4x+3} dx$$
$$I_1 = 10 \int \frac{2x+4}{x^2+4x+3} dx - 25 \int \frac{dx}{x^2+4x+3}$$
$$I_1 = 10 \log|x^2+4x+3| - 25 \int \frac{dx}{(x+2)^2 - 1^2}$$
$$I_1 = 10 \log|x^2+4x+3| - 25 \cdot \frac{1}{2} \log|\frac{x+2-1}{x+2+1}|$$
Applying limits 1 to 2 for \(I_1\):
$$I_1 = \left[ 10 \log|x^2+4x+3| - \frac{25}{2} \log|\frac{x+1}{x+3}| \right]_{1}^{2}$$
$$I_1 = 10(\log 15 - \log 8) - \frac{25}{2}(\log \frac{3}{5} - \log \frac{2}{4})$$
$$I_1 = 10 \log 15 - 10 \log 8 - \frac{25}{2} \log 3 + \frac{25}{2} \log 5 + \frac{25}{2} \log \frac{1}{2}$$
Simplifying this expression results in \(I_1 = 10 \log 5 + 10 \log 3 - 30 \log 2 - \frac{25}{2} \log 3 + \frac{25}{2} \log 5 - \frac{25}{2} \log 2\)
Alternatively, substituting back into \(I = 5 - I_1\).
Final simplified form: \(5 - [ 10 \log(\frac{15}{8}) - \frac{25}{2} \log(\frac{6}{5}) ]\)
Question 17. Evaluate \(\int_{0}^{\pi/4} (2\sec^2 x + x^3 + 2) dx\)
$$I = \left[ 2 \tan x + \frac{x^4}{4} + 2x \right]_{0}^{\pi/4}$$
$$I = \left( 2 \tan \frac{\pi}{4} + \frac{(\pi/4)^4}{4} + 2(\frac{\pi}{4}) \right) - (0)$$
$$I = 2(1) + \frac{\pi^4}{1024} + \frac{\pi}{2}$$
$$I = 2 + \frac{\pi}{2} + \frac{\pi^4}{1024}$$
Question 18. Evaluate \(\int_{0}^{\pi} (\sin^2 \frac{x}{2} - \cos^2 \frac{x}{2}) dx\)
Use formula: \(\cos^2 A - \sin^2 A = \cos 2A\).
$$\sin^2 \frac{x}{2} - \cos^2 \frac{x}{2} = -(\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}) = -\cos x$$
$$I = \int_{0}^{\pi} (-\cos x) dx = -[\sin x]_{0}^{\pi}$$
$$I = -(\sin \pi - \sin 0) = -(0 - 0) = 0$$
Answer: 0
Question 19. Evaluate \(\int_{0}^{2} \frac{6x+3}{x^2+4} dx\)
Split the integral:
$$I = \int_{0}^{2} \frac{6x}{x^2+4} dx + \int_{0}^{2} \frac{3}{x^2+4} dx$$
$$I = 3 \int_{0}^{2} \frac{2x}{x^2+4} dx + 3 \int_{0}^{2} \frac{dx}{x^2+2^2}$$
$$I = 3 [\log(x^2+4)]_{0}^{2} + 3 \cdot \frac{1}{2} [\tan^{-1} \frac{x}{2}]_{0}^{2}$$
$$I = 3 (\log 8 - \log 4) + \frac{3}{2} (\tan^{-1} 1 - \tan^{-1} 0)$$
$$I = 3 \log 2 + \frac{3}{2} (\frac{\pi}{4}) = 3 \log 2 + \frac{3\pi}{8}$$
Answer: \(3 \log 2 + \frac{3\pi}{8}\)
Question 20. Evaluate \(\int_{0}^{1} (x e^x + \sin \frac{\pi x}{4}) dx\)
Part 1: \(\int x e^x dx\). Using integration by parts: \(x e^x - \int e^x dx = x e^x - e^x\).
$$[x e^x - e^x]_{0}^{1} = (e - e) - (0 - 1) = 1$$
Part 2: \(\int \sin \frac{\pi x}{4} dx = \frac{-\cos(\pi x / 4)}{\pi / 4} = -\frac{4}{\pi} \cos \frac{\pi x}{4}\).
$$\left[ -\frac{4}{\pi} \cos \frac{\pi x}{4} \right]_{0}^{1} = -\frac{4}{\pi} (\cos \frac{\pi}{4} - \cos 0)$$
$$= -\frac{4}{\pi} (\frac{1}{\sqrt{2}} - 1) = \frac{4}{\pi} - \frac{2\sqrt{2}}{\pi}$$
Total \(I = 1 + \frac{4 - 2\sqrt{2}}{\pi}\)
Question 21. \(\int_{1}^{\sqrt{3}} \frac{dx}{1+x^2}\) equals
$$I = [\tan^{-1} x]_{1}^{\sqrt{3}}$$
$$I = \tan^{-1} \sqrt{3} - \tan^{-1} 1$$
$$I = \frac{\pi}{3} - \frac{\pi}{4} = \frac{4\pi - 3\pi}{12} = \frac{\pi}{12}$$
A. \(\pi/3\)
B. \(2\pi/3\)
C. \(\pi/6\)
D. \(\pi/12\)
Correct Option: D
Question 22. \(\int_{0}^{2/3} \frac{dx}{4+9x^2}\) equals
$$I = \int_{0}^{2/3} \frac{dx}{(2)^2 + (3x)^2}$$
Using \(\int \frac{dx}{a^2 + (bx)^2} = \frac{1}{b} \cdot \frac{1}{a} \tan^{-1} \frac{bx}{a}\)
$$I = \frac{1}{3} \cdot \frac{1}{2} [\tan^{-1} \frac{3x}{2}]_{0}^{2/3}$$
$$I = \frac{1}{6} [\tan^{-1}(\frac{3}{2} \cdot \frac{2}{3}) - \tan^{-1} 0]$$
$$I = \frac{1}{6} (\tan^{-1} 1 - 0) = \frac{1}{6} \cdot \frac{\pi}{4} = \frac{\pi}{24}$$
A. \(\pi/6\)
B. \(\pi/12\)
C. \(\pi/24\)
D. \(\pi/4\)
Correct Option: C