AP Inter 2nd Year Maths Exercise 7h Solutions

Referring to the AP Inter 2nd Year Maths Study Material Chapter 7 Integrals Exercise 7h Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Integrals Solutions Exercise 7h

I.

Question 1.
Evaluate \(\int_{-1}^1(x+1) d x\)
Solution:
Let I = \(\int_{-1}^1(x+1) d x=\left[\frac{x^2}{2}+x\right]_{-1}^1=\left[\frac{1}{2}+1\right]-\left[\frac{1}{2}-1\right]=\frac{1}{2}+1-\frac{1}{2}+1=2\)

Question 2.
Evaluate \(\int_2^3 \frac{1}{x} d x\)
Solution:
Let I = \(\int_2^3 \frac{1}{x} d x=[\log |x|]_2^3=\log |3|-\log |2|=\log \frac{3}{2}\)

Question 3.
Evaluate \(\int_1^2\left(4 x^3-5 x^2+6 x+9\right) d x\)
Solution:
\(\int_1^2\left[4 x^3-5 x^2+6 x+9\right] d x=\left[4 \frac{x^4}{4}-5 \frac{x^3}{3}+6 \frac{x^2}{2}+9 x\right]_1^2\)
= \(\left[x^4-\frac{5}{3} x^3+3 x^2+9 x\right]_1^2=\left[2^4-\frac{5}{3}(2)^3+3(2)^2+9(2)\right]-\left[1-\frac{5}{3}+3+9\right]\)
= \(\left[16-\frac{40}{3}+12+18\right]-\left[13-\frac{5}{3}\right]=\left[46-\frac{40}{3}\right]-\left[13-\frac{5}{3}\right]\)
= \(46-\frac{40}{3}-13+\frac{5}{3}=33-\frac{40}{3}+\frac{5}{3}=\frac{99-40+5}{3}=\frac{104-40}{3}=\frac{64}{3}\).

Question 4.
Evaluate \(\int_0^\pi 4 \sin 2 x d x\)
Solution:
\(\int_0^{\frac{\pi}{4}} \sin 2 x d x=\left[\frac{-\cos 2 x}{2}\right]_0^{\frac{\pi}{4}}=\left[\frac{-\cos 2 \frac{\pi}{4}}{2}\right]-\left[\frac{-\cos 0}{2}\right]=0-\left(\frac{-1}{2}\right)=0+\frac{1}{2}=\frac{1}{2}\) [∵ \(\cos \frac{\pi}{2}\) = 0]

Question 5.
Evaluate \(\int_0^\pi 2 \cos 2 x d x\)
Solution:
\(\int_0^{\frac{\pi}{2}} \cos 2 x d x=\left[\frac{\sin 2 x}{2}\right]_0^{\frac{\pi}{2}}=\frac{\sin \pi}{2}-\frac{\sin 0}{2}=\frac{0}{2}-\frac{0}{2}=0\) [∵ sin π = sin 180° = 0]

Question 6.
Evaluate \(\int_4^5 e^x d x\)
Solution:
Let I = \(\int_4^5 e^x d x=\left[e^x\right]_4^5\) = e⁵ – e⁴ = e⁴(e – 1)

Question 7.
Evaluate \(\int_0^\pi 4 \tan x d x\)
Solution:
Let I = \(\int_0^{\frac{\pi}{4}} \tan x d x=[-\log \cos x]_0^{\pi / 4}=-\log \left|\cos \frac{\pi}{4}\right|+\log |\cos 0|\)
= \(-\log \left|\frac{1}{\sqrt{2}}\right|+\log |1|=-\log (2)^{\frac{-1}{2}}=\frac{1}{2} \log 2\)

Question 8.
Evaluate \(\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} {cosec} x d x\)
Solution:
I = \(\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} {cosec} x d x=[\log |{cosec} x-\cot x|]_{\pi / 6}^{\pi / 4}=\log \left|{cosec} \frac{\pi}{4}-\cot \frac{\pi}{4}\right|-\log \left|{cosec} \frac{\pi}{6}-\cot \frac{\pi}{6}\right|\)
= \(\log |\sqrt{2}-1|-\log |2-\sqrt{3}|=\log \left(\frac{\sqrt{2}-1}{2-\sqrt{3}}\right)\)

Question 9.
Evaluate \(\int_0^1 \frac{d x}{\sqrt{1-x^2}}\)
Solution:
Let I = \(\int_0^1 \frac{d x}{\sqrt{1-x^2}}=\left[\sin ^{-1} x\right]_0^1=\sin ^{-1}(1)-\sin ^{-1}(0)=\frac{\pi}{2}-0=\frac{\pi}{2}\)

Question 10.
Evaluate \(\int_0^1 \frac{d x}{1+x^2}\)
Solution:
Let I = \(\int_0^1 \frac{d x}{1+x^2}=\left[\tan ^{-1} x\right]_0^1=\tan ^{-1}(1)-\tan ^{-1}(0)=\frac{\pi}{4}\)

Question 11.
Evaluate \(\int_2^3 \frac{d x}{x^2-1}\)
Solution:
Let I = \(\int_2^3 \frac{\mathrm{dx}}{\mathrm{x}^2-1}=\left[\frac{1}{2} \log \left|\frac{\mathrm{x}-1}{\mathrm{x}+1}\right|\right]_2^3\)
= \(\frac{1}{2}\left[\log \left|\frac{3-1}{3+1}\right|-\log \left|\frac{2-1}{2+1}\right|\right]=\frac{1}{2}\left[\log \left|\frac{2}{4}\right|-\log \left|\frac{1}{3}\right|\right]=\frac{1}{2}\left[\log \frac{1}{2}-\log \frac{1}{3}\right]=\frac{1}{2}\left[\log \frac{3}{2}\right]\)

Question 12.
Evaluate \(\int_2^3 \frac{x d x}{x^2+1}\)
Solution:
Let I = \(\int_2^3 \frac{x}{x^2+1} d x=\frac{1}{2} \int_2^3 \frac{2 x}{x^2+1} d x=\frac{1}{2}\left[\log \left(1+x^2\right)\right]_2^3\)
= \(\frac{1}{2}\left[\log \left(1+3^2\right)-\log \left(1+2^2\right)\right]=\frac{1}{2}[\log (10)-\log (5)]=\frac{1}{2} \log \left(\frac{10}{5}\right)=\frac{1}{2} \log 2\)

Question 13.
Evaluate \(\int_0^1 x e^{x^2} d x\)
Solution:
Let I = \(\int_0^1 x e^{x^2} d x\) put x² = t ⇒ 2x dx = dt
As x → 0, t → 0 and as x → 1, t → 1
∴ I = \(\frac{1}{2} \int_0^1 e^t d t=\frac{1}{2}\left[e^t\right]_0^1=\frac{1}{2} e-\frac{1}{2} e^0=\frac{1}{2}(e-1)\)

Question 14.
Evaluate \(\int_0^{\pi / 4}\left(2 \sec ^2 x+x^3+2\right) d x\)
Solution:
\(\int_0^{\pi / 4}\left(2 \sec ^2 x+x^3+2\right) d x=\left[2 \tan x+\frac{x^4}{4}+2 x\right]_0^{\pi / 4}\)
= \(\left[\left(2 \tan \frac{\pi}{4}+\frac{1}{4}\left(\frac{\pi}{4}\right)^4+2\left(\frac{\pi}{4}\right)\right)-(2 \tan 0+0+0)\right]=2+\frac{\pi^4}{4^5}+\frac{\pi}{2}=2+\frac{\pi}{2}+\frac{\pi^4}{1024}\)

Question 15.
Evaluate \(\int_0^\pi\left(\sin ^2 \frac{x}{2}-\cos ^2 \frac{x}{2}\right) d x\)
Solution:
Let I = \(\int_0^\pi\left(\sin ^2 \frac{x}{2}-\cos ^2 \frac{x}{2}\right) d x=-\int_0^\pi\left(\cos ^2 \frac{x}{2}-\sin ^2 \frac{x}{2}\right) d x\)
= \(-\int_0^\pi \cos x d x=-[\sin x]_0^\pi\) = -(sin π – sin 0) = 0 – 0 = 0

Question 16.
Evaluate \(\int_0^{\frac{\pi}{2}} \cos ^2 x d x\)
Solution:
Let I = \(\int_0^{\frac{\pi}{2}} \cos ^2 x d x=\int_0^{\pi / 2}\left(\frac{1+\cos 2 x}{2}\right) d x=\left[\frac{x}{2}+\frac{\sin 2 x}{4}\right]_0^{\pi / 2}=\frac{1}{2}\left[x+\frac{\sin 2 x}{2}\right]_0^{\pi / 2}\)
= \(\frac{1}{2}\left[\left(\frac{\pi}{2}+\frac{\sin \pi}{2}\right)-\left(0+\frac{\sin 0}{2}\right)\right]=\frac{1}{2}\left[\frac{\pi}{2}+0-0-0\right]=\frac{\pi}{4}\)

Question 17.
Find the integral of \(\int_0^1 \frac{\mathrm{dx}}{\sqrt{1+\mathrm{x}}-\sqrt{\mathrm{x}}}\)
Solution:

Question 18.
Evaluate \(\int_0^{\frac{\pi}{2}} \sin ^3 x d x\)
Solution:
Let I = \(\int_0^{\frac{\pi}{2}} \sin ^3 x d x\)
I = \(\int_0^{\frac{\pi}{2}} \sin ^2 x \cdot \sin x d x=\int_0^{\frac{\pi}{2}}\left(1-\cos ^2 x\right) \sin x d x=\int_0^{\frac{\pi}{2}} \sin x d x-\int_0^{\frac{\pi}{2}} \cos ^2 x \cdot \sin x d x\)
= \([-\cos x]_0^{\frac{\pi}{2}}+\left[\frac{\cos ^3 x}{3}\right]_0^{\frac{\pi}{2}}=\left[\cos \frac{\pi}{2}-\cos 0\right]+\frac{1}{3}\left[\cos ^3 \frac{\pi}{2}-\cos ^3 0\right]=1+\frac{1}{3}[-1]=1-\frac{1}{3}=\frac{2}{3}\)

II.

Question 1.
Evaluate \(\int_0^1 \frac{2 x+3}{5 x^2+1} d x\)
Solution:

Question 2.
Evaluate \(\int_0^2 \frac{6 x+3}{x^2+4} d x\)
Solution:

Question 3.
Evaluate \(\int_1^2 \frac{5 x^2}{x^2+4 x+3}\)
Solution:
Let I = \(\int_1^2 \frac{5 x^2}{x^2+4 x+3}\) dx
Dividing the Nr. & Dr. By x² + 4x + 3 and simplifying, we get
Dividing 5x² by x² + 4x + 3, we get
I = \(\int_1^2\left[5-\frac{20 x+15}{x^2+4 x+3}\right] d x=\int_1^2 5 d x-\int_1^2 \frac{20 x+15}{x^2+4 x+3} d x=[5 x]_1^2-\int_1^2 \frac{20 x+15}{x^2+4 x+3} d x\) …(1)
Let I₁ = \(\int_1^2 \frac{20 x+15}{x^2+4 x+3} d x\) …………(1)
Let 20x + 15 = A\(\frac{d}{d x}\)(x² + 4x + 3) + B = 2Ax + (4A + B)
Equating the coefficients of x and constant term, we get
A = 10 and B = -25
∴ 20x + 15 = 10(2x + 4) – 25 [∵ \(\int \frac{d x}{x^2-a^2}=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C\)]
⇒ I₁ = \(10 \int_1^2 \frac{2 x+4}{x^2+4 x+3} d x-25 \int_1^2 \frac{d x}{(x+2)^2-1^2}=10\left[\log \left|x^2+4 x+3\right|\right]_1^2-25\left[\frac{1}{2} \log \left(\frac{x+2-1}{x+2+1}\right)\right]_1^2\)
= \(\left[10 \log \left(x^2+4 x+3\right)\right]_1^2-25\left[\frac{1}{2} \log \left(\frac{x+1}{x+3}\right)\right]_1^2\)
= [10log15 – 10log8] – 25 \(\left[\frac{1}{2} \log \frac{3}{5}-\frac{1}{2} \log \frac{2}{4}\right]\)
= [10 log(5 × 3) – 10 log(4 × 2)] – \(\frac{25}{2}\)[log 3 – log 5 – log 2 + log 4]
= [10 log5 + 10log3 – 10log4 – 10log2] – \(\frac{25}{2}\)[log3 – log5 – log2 + log4]
= \(\left[10+\frac{25}{2}\right] \log 5+\left[-10-\frac{25}{2}\right] \log 4+\left[10-\frac{25}{2}\right] \log 3+\left[-10+\frac{25}{2}\right] \log 2\)
= \(\frac{45}{2} \log 5-\frac{45}{2} \log 4-\frac{5}{2} \log 3+\frac{5}{2} \log 2=\frac{45}{2} \log \frac{5}{4}-\frac{5}{2} \log \frac{3}{2}\)
Substituting the value I₁, in (1), we get
I = \(5-\left[\frac{45}{2} \log \frac{5}{4}-\frac{5}{2} \log \frac{3}{2}\right]=5-\frac{5}{2}\left[9 \log \frac{5}{4}-\log \frac{3}{2}\right]\)

Question 4.
Evaluate \(\int_0^1\left[x e^x+\sin \frac{\pi x}{4}\right] d x\)
Solution:
Let I = \(\int_0^1\left[x e^x+\sin \frac{\pi x}{4}\right] d x\)

Question 5.
Evaluate \(\int_1^3 \frac{d x}{x^2(x+1)}\)
Solution:
Let, \(\frac{1}{x^2(x+1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1} \Rightarrow \frac{A x(x+1)+B(x+1)+C\left(x^2\right)}{x\left(x^2\right)(x+1)}\)
⇒ 1 = Ax(x + 1) + B(x + 1) + C(x²) ⇒ 1 = Ax² + Ax + Bx + B + Cx²
Equatingthe coefficients of x², x and constant terms, we get A + C = 0, A + B = 0, B = 1
On solving these equations, we get A = -1, C = 1, B = 1
∴ \(\frac{1}{x^2(x+1)}=\frac{-1}{x}+\frac{1}{x^2}+\frac{1}{(x+1)}\)
⇒ I = \(\int_1^3\left[-\frac{1}{x}+\frac{1}{x^2}+\frac{1}{(x+1)}\right] d x=\left[-\log x-\frac{1}{x}+\log (x+1)\right]_1^3\)
= \(\left[\log \left(\frac{x+1}{x}\right)-\frac{1}{x}\right]_1^3=\log \left(\frac{4}{3}\right)-\frac{1}{3}-\log \left(\frac{2}{1}\right)+1\)
= log 4 – log 3 – log 2 + \(\frac{2}{3}\) = log 2 – log 3 + \(\frac{2}{3}\) = \(\log \left(\frac{2}{3}\right)+\frac{2}{3}\), Hence proved.

Question 6.
Evaluate \(\int_0^1 \frac{x^{\frac{1}{4}}}{1+x^{\frac{1}{2}}} d x\)
Solution:
Put x^1/4 = t ⇒ x = t⁴ ⇒ dx = 4t³ dt.
Also, when x = 0 we get t = 0 and when x = 1 we get t = 1^1/4 = 1