86450
\(\int_{2016}^{2017} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{4033-x}} d x\) is equal to
1 \(1 / 4\)
2 \(3 / 2\)
3 \(2017 / 2\)
4 \(1 / 2\)
5 508
Explanation:
(D) : Given, \(\int_{2016}^{2017} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{4033-x}} d x \tag{i}\) \(\because \quad \int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\) \(I=\int_{2016}^{2017} \frac{\sqrt{4033-x}}{\sqrt{4033-x}+\sqrt{x}} d x\) Adding equation (i) and (ii), we get- \(2 I=\int_{2016}^{2017} \frac{\sqrt{4033-x}+\sqrt{x}}{\sqrt{4033-x}+\sqrt{x}} d x\) \(2 I=\int_{2016}^{2017} d x \Rightarrow I=\frac{1}{2}[x]_{2016}^{2017}\) \(I=\frac{1}{2}[2017-2016]=\frac{1}{2}\)
Kerala CEE-2017
Integral Calculus
86451
\(\int_{0}^{\pi / 2} \log \left(\frac{\cos x}{\sin x}\right) d x\) is equal to
86450
\(\int_{2016}^{2017} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{4033-x}} d x\) is equal to
1 \(1 / 4\)
2 \(3 / 2\)
3 \(2017 / 2\)
4 \(1 / 2\)
5 508
Explanation:
(D) : Given, \(\int_{2016}^{2017} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{4033-x}} d x \tag{i}\) \(\because \quad \int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\) \(I=\int_{2016}^{2017} \frac{\sqrt{4033-x}}{\sqrt{4033-x}+\sqrt{x}} d x\) Adding equation (i) and (ii), we get- \(2 I=\int_{2016}^{2017} \frac{\sqrt{4033-x}+\sqrt{x}}{\sqrt{4033-x}+\sqrt{x}} d x\) \(2 I=\int_{2016}^{2017} d x \Rightarrow I=\frac{1}{2}[x]_{2016}^{2017}\) \(I=\frac{1}{2}[2017-2016]=\frac{1}{2}\)
Kerala CEE-2017
Integral Calculus
86451
\(\int_{0}^{\pi / 2} \log \left(\frac{\cos x}{\sin x}\right) d x\) is equal to
86450
\(\int_{2016}^{2017} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{4033-x}} d x\) is equal to
1 \(1 / 4\)
2 \(3 / 2\)
3 \(2017 / 2\)
4 \(1 / 2\)
5 508
Explanation:
(D) : Given, \(\int_{2016}^{2017} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{4033-x}} d x \tag{i}\) \(\because \quad \int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\) \(I=\int_{2016}^{2017} \frac{\sqrt{4033-x}}{\sqrt{4033-x}+\sqrt{x}} d x\) Adding equation (i) and (ii), we get- \(2 I=\int_{2016}^{2017} \frac{\sqrt{4033-x}+\sqrt{x}}{\sqrt{4033-x}+\sqrt{x}} d x\) \(2 I=\int_{2016}^{2017} d x \Rightarrow I=\frac{1}{2}[x]_{2016}^{2017}\) \(I=\frac{1}{2}[2017-2016]=\frac{1}{2}\)
Kerala CEE-2017
Integral Calculus
86451
\(\int_{0}^{\pi / 2} \log \left(\frac{\cos x}{\sin x}\right) d x\) is equal to
NEET Test Series from KOTA - 10 Papers In MS WORD
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Integral Calculus
86450
\(\int_{2016}^{2017} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{4033-x}} d x\) is equal to
1 \(1 / 4\)
2 \(3 / 2\)
3 \(2017 / 2\)
4 \(1 / 2\)
5 508
Explanation:
(D) : Given, \(\int_{2016}^{2017} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{4033-x}} d x \tag{i}\) \(\because \quad \int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\) \(I=\int_{2016}^{2017} \frac{\sqrt{4033-x}}{\sqrt{4033-x}+\sqrt{x}} d x\) Adding equation (i) and (ii), we get- \(2 I=\int_{2016}^{2017} \frac{\sqrt{4033-x}+\sqrt{x}}{\sqrt{4033-x}+\sqrt{x}} d x\) \(2 I=\int_{2016}^{2017} d x \Rightarrow I=\frac{1}{2}[x]_{2016}^{2017}\) \(I=\frac{1}{2}[2017-2016]=\frac{1}{2}\)
Kerala CEE-2017
Integral Calculus
86451
\(\int_{0}^{\pi / 2} \log \left(\frac{\cos x}{\sin x}\right) d x\) is equal to