86454
\(\int_{0}^{10} \frac{x^{10}}{(10-x)^{10}+x^{10}} d x\) is equal to
1 10
2 5
3 2
4 \(\frac{1}{2}\)
5 0
Explanation:
(B) : Given,\(I=\int_{0}^{10} \frac{x^{10}}{(10-x)^{10}+x^{10}} d x\) \(\because \quad \int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\) \(I=\int_{0}^{10} \frac{(10-x)^{10}}{x^{10}+(10-x)^{10}} d x\) Adding equation (i) and (ii), we get- \(2 I=\int_{0}^{10} \frac{(10-x)^{10}+x^{10}}{(10-x)^{10}+x^{10}} d x\) \(2 I=\int_{0}^{10} 1 d x\) \(2 I=[x]_{0}^{10}\) \(I=\frac{1}{2}[10-0]=5\)
86456
If \(\int_{0}^{\pi} \mathrm{x} f(\sin \mathrm{x}) \mathrm{d} \mathrm{x}=\mathrm{A} \int_{0}^{\pi / 2} f(\sin \mathrm{x}) \mathrm{d} \mathrm{x}\), then \(\mathrm{A}\) is equal to
1 0
2 \(\pi\)
3 \(\frac{\pi}{4}\)
4 \(2 \pi\)
5 \(3 \pi\)
Explanation:
(B) : Given, \(\int_{0}^{\pi} x f(\sin x) d x=A \int_{0}^{\pi / 2} f(\sin x) d x \tag{i}\) Let, \(\quad \mathrm{I}=\int_{0}^{\pi} x f(\sin \mathrm{x}) \mathrm{dx}\) \(=\int_{0}^{\pi}(\pi-\mathrm{x}) \mathrm{f}(\sin (\pi-\mathrm{x})) \mathrm{dx}\) \(\left(\because \int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\right)\) \(I=\int_{0}^{\pi}(\pi-x) f(\sin x) d x\) On adding equation (i )and (ii), we get - \(2 I=\pi \int_{0}^{\pi} f(\sin x) d x\) \(\quad\left(\because \int_{0}^{a} f(x) d x=2 \int_{0}^{a / 2} f(x) d x\right)\) \(I=\frac{\pi}{2} \int_{0}^{\pi} \mathrm{f}(\sin \mathrm{x}) \mathrm{dx}=\pi \int_{0}^{\pi / 2} \mathrm{f}(\sin \mathrm{x}) \mathrm{dx}\) \(A \int_{0}^{\pi / 2} f(\sin x) d x=\pi \int_{0}^{\pi / 2} f(\sin x) d x\) \(\mathrm{A}=\pi\)
Kerala CEE-2008
Integral Calculus
86457
\(\int_{-1}^{1} \frac{17 x^{5}-x^{4}+29 x^{3}-31 x+1}{x^{2}+1} d x\) is
1 \(4 / 5\)
2 \(5 / 4\)
3 \(4 / 3\)
4 \(3 / 4\)
5 6
Explanation:
(C) : Given, \(I =\int_{-1}^{1} \frac{17 x^{5}-x^{4}+29 x^{3}-31 x+1}{x^{2}+1} d x\) \(=\int_{-1}^{1}\left(\frac{17 x^{5}+29 x^{3}-31 x}{x^{2}+1}-\frac{x^{4}-1}{x^{2}+1}\right) d x\) \(=\int_{-1}^{1} \frac{17 x^{5}+29 x^{3}-31 x}{x^{2}+1} d x-\int_{-1}^{1} \frac{\left(x^{2}+1\right)\left(x^{2}-1\right)}{x^{2}+1} d x\) \(=\int_{-1}^{1} \frac{17 x^{5}+29 x^{3}-31 x}{x^{2}+1} d x-\int_{-1}^{1}\left(x^{2}-1\right) d x\) \(=0-2 \int_{0}^{1} \frac{\left(\mathrm{x}^{2}-1\right)\left(\mathrm{x}^{2}+1\right)}{\left(\mathrm{x}^{2}+1\right)} \mathrm{dx}\) \(=-2\left[\left(\frac{x^{3}}{3}-x\right)\right]_{0}^{1}=-2\left[\frac{1}{3}-1\right]=\frac{4}{3}\)
86454
\(\int_{0}^{10} \frac{x^{10}}{(10-x)^{10}+x^{10}} d x\) is equal to
1 10
2 5
3 2
4 \(\frac{1}{2}\)
5 0
Explanation:
(B) : Given,\(I=\int_{0}^{10} \frac{x^{10}}{(10-x)^{10}+x^{10}} d x\) \(\because \quad \int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\) \(I=\int_{0}^{10} \frac{(10-x)^{10}}{x^{10}+(10-x)^{10}} d x\) Adding equation (i) and (ii), we get- \(2 I=\int_{0}^{10} \frac{(10-x)^{10}+x^{10}}{(10-x)^{10}+x^{10}} d x\) \(2 I=\int_{0}^{10} 1 d x\) \(2 I=[x]_{0}^{10}\) \(I=\frac{1}{2}[10-0]=5\)
86456
If \(\int_{0}^{\pi} \mathrm{x} f(\sin \mathrm{x}) \mathrm{d} \mathrm{x}=\mathrm{A} \int_{0}^{\pi / 2} f(\sin \mathrm{x}) \mathrm{d} \mathrm{x}\), then \(\mathrm{A}\) is equal to
1 0
2 \(\pi\)
3 \(\frac{\pi}{4}\)
4 \(2 \pi\)
5 \(3 \pi\)
Explanation:
(B) : Given, \(\int_{0}^{\pi} x f(\sin x) d x=A \int_{0}^{\pi / 2} f(\sin x) d x \tag{i}\) Let, \(\quad \mathrm{I}=\int_{0}^{\pi} x f(\sin \mathrm{x}) \mathrm{dx}\) \(=\int_{0}^{\pi}(\pi-\mathrm{x}) \mathrm{f}(\sin (\pi-\mathrm{x})) \mathrm{dx}\) \(\left(\because \int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\right)\) \(I=\int_{0}^{\pi}(\pi-x) f(\sin x) d x\) On adding equation (i )and (ii), we get - \(2 I=\pi \int_{0}^{\pi} f(\sin x) d x\) \(\quad\left(\because \int_{0}^{a} f(x) d x=2 \int_{0}^{a / 2} f(x) d x\right)\) \(I=\frac{\pi}{2} \int_{0}^{\pi} \mathrm{f}(\sin \mathrm{x}) \mathrm{dx}=\pi \int_{0}^{\pi / 2} \mathrm{f}(\sin \mathrm{x}) \mathrm{dx}\) \(A \int_{0}^{\pi / 2} f(\sin x) d x=\pi \int_{0}^{\pi / 2} f(\sin x) d x\) \(\mathrm{A}=\pi\)
Kerala CEE-2008
Integral Calculus
86457
\(\int_{-1}^{1} \frac{17 x^{5}-x^{4}+29 x^{3}-31 x+1}{x^{2}+1} d x\) is
1 \(4 / 5\)
2 \(5 / 4\)
3 \(4 / 3\)
4 \(3 / 4\)
5 6
Explanation:
(C) : Given, \(I =\int_{-1}^{1} \frac{17 x^{5}-x^{4}+29 x^{3}-31 x+1}{x^{2}+1} d x\) \(=\int_{-1}^{1}\left(\frac{17 x^{5}+29 x^{3}-31 x}{x^{2}+1}-\frac{x^{4}-1}{x^{2}+1}\right) d x\) \(=\int_{-1}^{1} \frac{17 x^{5}+29 x^{3}-31 x}{x^{2}+1} d x-\int_{-1}^{1} \frac{\left(x^{2}+1\right)\left(x^{2}-1\right)}{x^{2}+1} d x\) \(=\int_{-1}^{1} \frac{17 x^{5}+29 x^{3}-31 x}{x^{2}+1} d x-\int_{-1}^{1}\left(x^{2}-1\right) d x\) \(=0-2 \int_{0}^{1} \frac{\left(\mathrm{x}^{2}-1\right)\left(\mathrm{x}^{2}+1\right)}{\left(\mathrm{x}^{2}+1\right)} \mathrm{dx}\) \(=-2\left[\left(\frac{x^{3}}{3}-x\right)\right]_{0}^{1}=-2\left[\frac{1}{3}-1\right]=\frac{4}{3}\)
86454
\(\int_{0}^{10} \frac{x^{10}}{(10-x)^{10}+x^{10}} d x\) is equal to
1 10
2 5
3 2
4 \(\frac{1}{2}\)
5 0
Explanation:
(B) : Given,\(I=\int_{0}^{10} \frac{x^{10}}{(10-x)^{10}+x^{10}} d x\) \(\because \quad \int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\) \(I=\int_{0}^{10} \frac{(10-x)^{10}}{x^{10}+(10-x)^{10}} d x\) Adding equation (i) and (ii), we get- \(2 I=\int_{0}^{10} \frac{(10-x)^{10}+x^{10}}{(10-x)^{10}+x^{10}} d x\) \(2 I=\int_{0}^{10} 1 d x\) \(2 I=[x]_{0}^{10}\) \(I=\frac{1}{2}[10-0]=5\)
86456
If \(\int_{0}^{\pi} \mathrm{x} f(\sin \mathrm{x}) \mathrm{d} \mathrm{x}=\mathrm{A} \int_{0}^{\pi / 2} f(\sin \mathrm{x}) \mathrm{d} \mathrm{x}\), then \(\mathrm{A}\) is equal to
1 0
2 \(\pi\)
3 \(\frac{\pi}{4}\)
4 \(2 \pi\)
5 \(3 \pi\)
Explanation:
(B) : Given, \(\int_{0}^{\pi} x f(\sin x) d x=A \int_{0}^{\pi / 2} f(\sin x) d x \tag{i}\) Let, \(\quad \mathrm{I}=\int_{0}^{\pi} x f(\sin \mathrm{x}) \mathrm{dx}\) \(=\int_{0}^{\pi}(\pi-\mathrm{x}) \mathrm{f}(\sin (\pi-\mathrm{x})) \mathrm{dx}\) \(\left(\because \int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\right)\) \(I=\int_{0}^{\pi}(\pi-x) f(\sin x) d x\) On adding equation (i )and (ii), we get - \(2 I=\pi \int_{0}^{\pi} f(\sin x) d x\) \(\quad\left(\because \int_{0}^{a} f(x) d x=2 \int_{0}^{a / 2} f(x) d x\right)\) \(I=\frac{\pi}{2} \int_{0}^{\pi} \mathrm{f}(\sin \mathrm{x}) \mathrm{dx}=\pi \int_{0}^{\pi / 2} \mathrm{f}(\sin \mathrm{x}) \mathrm{dx}\) \(A \int_{0}^{\pi / 2} f(\sin x) d x=\pi \int_{0}^{\pi / 2} f(\sin x) d x\) \(\mathrm{A}=\pi\)
Kerala CEE-2008
Integral Calculus
86457
\(\int_{-1}^{1} \frac{17 x^{5}-x^{4}+29 x^{3}-31 x+1}{x^{2}+1} d x\) is
1 \(4 / 5\)
2 \(5 / 4\)
3 \(4 / 3\)
4 \(3 / 4\)
5 6
Explanation:
(C) : Given, \(I =\int_{-1}^{1} \frac{17 x^{5}-x^{4}+29 x^{3}-31 x+1}{x^{2}+1} d x\) \(=\int_{-1}^{1}\left(\frac{17 x^{5}+29 x^{3}-31 x}{x^{2}+1}-\frac{x^{4}-1}{x^{2}+1}\right) d x\) \(=\int_{-1}^{1} \frac{17 x^{5}+29 x^{3}-31 x}{x^{2}+1} d x-\int_{-1}^{1} \frac{\left(x^{2}+1\right)\left(x^{2}-1\right)}{x^{2}+1} d x\) \(=\int_{-1}^{1} \frac{17 x^{5}+29 x^{3}-31 x}{x^{2}+1} d x-\int_{-1}^{1}\left(x^{2}-1\right) d x\) \(=0-2 \int_{0}^{1} \frac{\left(\mathrm{x}^{2}-1\right)\left(\mathrm{x}^{2}+1\right)}{\left(\mathrm{x}^{2}+1\right)} \mathrm{dx}\) \(=-2\left[\left(\frac{x^{3}}{3}-x\right)\right]_{0}^{1}=-2\left[\frac{1}{3}-1\right]=\frac{4}{3}\)
86454
\(\int_{0}^{10} \frac{x^{10}}{(10-x)^{10}+x^{10}} d x\) is equal to
1 10
2 5
3 2
4 \(\frac{1}{2}\)
5 0
Explanation:
(B) : Given,\(I=\int_{0}^{10} \frac{x^{10}}{(10-x)^{10}+x^{10}} d x\) \(\because \quad \int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\) \(I=\int_{0}^{10} \frac{(10-x)^{10}}{x^{10}+(10-x)^{10}} d x\) Adding equation (i) and (ii), we get- \(2 I=\int_{0}^{10} \frac{(10-x)^{10}+x^{10}}{(10-x)^{10}+x^{10}} d x\) \(2 I=\int_{0}^{10} 1 d x\) \(2 I=[x]_{0}^{10}\) \(I=\frac{1}{2}[10-0]=5\)
86456
If \(\int_{0}^{\pi} \mathrm{x} f(\sin \mathrm{x}) \mathrm{d} \mathrm{x}=\mathrm{A} \int_{0}^{\pi / 2} f(\sin \mathrm{x}) \mathrm{d} \mathrm{x}\), then \(\mathrm{A}\) is equal to
1 0
2 \(\pi\)
3 \(\frac{\pi}{4}\)
4 \(2 \pi\)
5 \(3 \pi\)
Explanation:
(B) : Given, \(\int_{0}^{\pi} x f(\sin x) d x=A \int_{0}^{\pi / 2} f(\sin x) d x \tag{i}\) Let, \(\quad \mathrm{I}=\int_{0}^{\pi} x f(\sin \mathrm{x}) \mathrm{dx}\) \(=\int_{0}^{\pi}(\pi-\mathrm{x}) \mathrm{f}(\sin (\pi-\mathrm{x})) \mathrm{dx}\) \(\left(\because \int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\right)\) \(I=\int_{0}^{\pi}(\pi-x) f(\sin x) d x\) On adding equation (i )and (ii), we get - \(2 I=\pi \int_{0}^{\pi} f(\sin x) d x\) \(\quad\left(\because \int_{0}^{a} f(x) d x=2 \int_{0}^{a / 2} f(x) d x\right)\) \(I=\frac{\pi}{2} \int_{0}^{\pi} \mathrm{f}(\sin \mathrm{x}) \mathrm{dx}=\pi \int_{0}^{\pi / 2} \mathrm{f}(\sin \mathrm{x}) \mathrm{dx}\) \(A \int_{0}^{\pi / 2} f(\sin x) d x=\pi \int_{0}^{\pi / 2} f(\sin x) d x\) \(\mathrm{A}=\pi\)
Kerala CEE-2008
Integral Calculus
86457
\(\int_{-1}^{1} \frac{17 x^{5}-x^{4}+29 x^{3}-31 x+1}{x^{2}+1} d x\) is
1 \(4 / 5\)
2 \(5 / 4\)
3 \(4 / 3\)
4 \(3 / 4\)
5 6
Explanation:
(C) : Given, \(I =\int_{-1}^{1} \frac{17 x^{5}-x^{4}+29 x^{3}-31 x+1}{x^{2}+1} d x\) \(=\int_{-1}^{1}\left(\frac{17 x^{5}+29 x^{3}-31 x}{x^{2}+1}-\frac{x^{4}-1}{x^{2}+1}\right) d x\) \(=\int_{-1}^{1} \frac{17 x^{5}+29 x^{3}-31 x}{x^{2}+1} d x-\int_{-1}^{1} \frac{\left(x^{2}+1\right)\left(x^{2}-1\right)}{x^{2}+1} d x\) \(=\int_{-1}^{1} \frac{17 x^{5}+29 x^{3}-31 x}{x^{2}+1} d x-\int_{-1}^{1}\left(x^{2}-1\right) d x\) \(=0-2 \int_{0}^{1} \frac{\left(\mathrm{x}^{2}-1\right)\left(\mathrm{x}^{2}+1\right)}{\left(\mathrm{x}^{2}+1\right)} \mathrm{dx}\) \(=-2\left[\left(\frac{x^{3}}{3}-x\right)\right]_{0}^{1}=-2\left[\frac{1}{3}-1\right]=\frac{4}{3}\)
