86528
The value of definite integral \(\int_{0}^{\frac{\pi}{2}} \log (\tan x) d x\) is
1 0
2 \(\frac{\pi}{4}\)
3 \(\frac{\pi}{2}\)
4 \(\pi\)
Explanation:
(A) : Let, \(= \int_{0}^{\frac{\pi}{2}} \log (\tan x) d x \) \(=\int_{0}^{\frac{\pi}{2}} \log \left\{\tan \left(\frac{\pi}{2}-x\right)\right\} d x\) \(=\int_{0}^{\frac{\pi}{2}} \log (\cot x) d x \tag{ii}\) On adding equation (i) and (ii), we get - \(\therefore 2 I=\int_{0}^{\frac{\pi}{2}} \log (\tan x) d x+\int_{0}^{\frac{\pi}{2}} \log (\cot x) d x\) \(2 \mathrm{I}=\int_{0}^{\frac{\pi}{2}}[\log \tan x+\log \cot x] d x\) \(2 I=\int_{0}^{\frac{\pi}{2}} \log (\tan x \cdot \cot x) d x \Rightarrow 2 I=\int_{0}^{\frac{\pi}{2}} \log (1) d x\) \(I = 0\)
BITSAT-2019
Integral Calculus
86529
The value of \(I=\int_{0}^{1} x\left|x-\frac{1}{2}\right| d x\) is
1 \(\frac{1}{3}\)
2 \(\frac{1}{4}\)
3 \(\frac{1}{8}\)
4 None of these
Explanation:
(C) : Given, \(I=\int_{0}^{1} x\left|x-\frac{1}{2}\right| d x\) \(=-\int_{0}^{1 / 2} x\left(x-\frac{1}{2}\right) d x+\int_{1 / 2}^{1} x\left(x-\frac{1}{2}\right) d x\) \(=\int_{0}^{1 / 2}\left(\frac{x}{2}-x^{2}\right) d x+\int_{1 / 2}^{1}\left(x^{2}-\frac{x}{2}\right) d x\) \(=\left[\frac{x^{2}}{4}-\frac{x^{3}}{3}\right]_{0}^{1 / 2}+\left[\frac{x^{3}}{3}-\frac{x^{2}}{4}\right]_{1 / 2}^{1}\) \(\left.=\left(\frac{1}{16}-\frac{1}{24}\right)+\left[\left(\frac{1}{3}-\frac{1}{4}\right)-\frac{1}{24}+\frac{1}{16}\right)\right]\) \(=\left(\frac{6-4}{96}\right)+\left(\frac{32-24-4+6}{96}\right)=\frac{12}{96}=\frac{1}{8}\)
86528
The value of definite integral \(\int_{0}^{\frac{\pi}{2}} \log (\tan x) d x\) is
1 0
2 \(\frac{\pi}{4}\)
3 \(\frac{\pi}{2}\)
4 \(\pi\)
Explanation:
(A) : Let, \(= \int_{0}^{\frac{\pi}{2}} \log (\tan x) d x \) \(=\int_{0}^{\frac{\pi}{2}} \log \left\{\tan \left(\frac{\pi}{2}-x\right)\right\} d x\) \(=\int_{0}^{\frac{\pi}{2}} \log (\cot x) d x \tag{ii}\) On adding equation (i) and (ii), we get - \(\therefore 2 I=\int_{0}^{\frac{\pi}{2}} \log (\tan x) d x+\int_{0}^{\frac{\pi}{2}} \log (\cot x) d x\) \(2 \mathrm{I}=\int_{0}^{\frac{\pi}{2}}[\log \tan x+\log \cot x] d x\) \(2 I=\int_{0}^{\frac{\pi}{2}} \log (\tan x \cdot \cot x) d x \Rightarrow 2 I=\int_{0}^{\frac{\pi}{2}} \log (1) d x\) \(I = 0\)
BITSAT-2019
Integral Calculus
86529
The value of \(I=\int_{0}^{1} x\left|x-\frac{1}{2}\right| d x\) is
1 \(\frac{1}{3}\)
2 \(\frac{1}{4}\)
3 \(\frac{1}{8}\)
4 None of these
Explanation:
(C) : Given, \(I=\int_{0}^{1} x\left|x-\frac{1}{2}\right| d x\) \(=-\int_{0}^{1 / 2} x\left(x-\frac{1}{2}\right) d x+\int_{1 / 2}^{1} x\left(x-\frac{1}{2}\right) d x\) \(=\int_{0}^{1 / 2}\left(\frac{x}{2}-x^{2}\right) d x+\int_{1 / 2}^{1}\left(x^{2}-\frac{x}{2}\right) d x\) \(=\left[\frac{x^{2}}{4}-\frac{x^{3}}{3}\right]_{0}^{1 / 2}+\left[\frac{x^{3}}{3}-\frac{x^{2}}{4}\right]_{1 / 2}^{1}\) \(\left.=\left(\frac{1}{16}-\frac{1}{24}\right)+\left[\left(\frac{1}{3}-\frac{1}{4}\right)-\frac{1}{24}+\frac{1}{16}\right)\right]\) \(=\left(\frac{6-4}{96}\right)+\left(\frac{32-24-4+6}{96}\right)=\frac{12}{96}=\frac{1}{8}\)
86528
The value of definite integral \(\int_{0}^{\frac{\pi}{2}} \log (\tan x) d x\) is
1 0
2 \(\frac{\pi}{4}\)
3 \(\frac{\pi}{2}\)
4 \(\pi\)
Explanation:
(A) : Let, \(= \int_{0}^{\frac{\pi}{2}} \log (\tan x) d x \) \(=\int_{0}^{\frac{\pi}{2}} \log \left\{\tan \left(\frac{\pi}{2}-x\right)\right\} d x\) \(=\int_{0}^{\frac{\pi}{2}} \log (\cot x) d x \tag{ii}\) On adding equation (i) and (ii), we get - \(\therefore 2 I=\int_{0}^{\frac{\pi}{2}} \log (\tan x) d x+\int_{0}^{\frac{\pi}{2}} \log (\cot x) d x\) \(2 \mathrm{I}=\int_{0}^{\frac{\pi}{2}}[\log \tan x+\log \cot x] d x\) \(2 I=\int_{0}^{\frac{\pi}{2}} \log (\tan x \cdot \cot x) d x \Rightarrow 2 I=\int_{0}^{\frac{\pi}{2}} \log (1) d x\) \(I = 0\)
BITSAT-2019
Integral Calculus
86529
The value of \(I=\int_{0}^{1} x\left|x-\frac{1}{2}\right| d x\) is
1 \(\frac{1}{3}\)
2 \(\frac{1}{4}\)
3 \(\frac{1}{8}\)
4 None of these
Explanation:
(C) : Given, \(I=\int_{0}^{1} x\left|x-\frac{1}{2}\right| d x\) \(=-\int_{0}^{1 / 2} x\left(x-\frac{1}{2}\right) d x+\int_{1 / 2}^{1} x\left(x-\frac{1}{2}\right) d x\) \(=\int_{0}^{1 / 2}\left(\frac{x}{2}-x^{2}\right) d x+\int_{1 / 2}^{1}\left(x^{2}-\frac{x}{2}\right) d x\) \(=\left[\frac{x^{2}}{4}-\frac{x^{3}}{3}\right]_{0}^{1 / 2}+\left[\frac{x^{3}}{3}-\frac{x^{2}}{4}\right]_{1 / 2}^{1}\) \(\left.=\left(\frac{1}{16}-\frac{1}{24}\right)+\left[\left(\frac{1}{3}-\frac{1}{4}\right)-\frac{1}{24}+\frac{1}{16}\right)\right]\) \(=\left(\frac{6-4}{96}\right)+\left(\frac{32-24-4+6}{96}\right)=\frac{12}{96}=\frac{1}{8}\)
86528
The value of definite integral \(\int_{0}^{\frac{\pi}{2}} \log (\tan x) d x\) is
1 0
2 \(\frac{\pi}{4}\)
3 \(\frac{\pi}{2}\)
4 \(\pi\)
Explanation:
(A) : Let, \(= \int_{0}^{\frac{\pi}{2}} \log (\tan x) d x \) \(=\int_{0}^{\frac{\pi}{2}} \log \left\{\tan \left(\frac{\pi}{2}-x\right)\right\} d x\) \(=\int_{0}^{\frac{\pi}{2}} \log (\cot x) d x \tag{ii}\) On adding equation (i) and (ii), we get - \(\therefore 2 I=\int_{0}^{\frac{\pi}{2}} \log (\tan x) d x+\int_{0}^{\frac{\pi}{2}} \log (\cot x) d x\) \(2 \mathrm{I}=\int_{0}^{\frac{\pi}{2}}[\log \tan x+\log \cot x] d x\) \(2 I=\int_{0}^{\frac{\pi}{2}} \log (\tan x \cdot \cot x) d x \Rightarrow 2 I=\int_{0}^{\frac{\pi}{2}} \log (1) d x\) \(I = 0\)
BITSAT-2019
Integral Calculus
86529
The value of \(I=\int_{0}^{1} x\left|x-\frac{1}{2}\right| d x\) is
1 \(\frac{1}{3}\)
2 \(\frac{1}{4}\)
3 \(\frac{1}{8}\)
4 None of these
Explanation:
(C) : Given, \(I=\int_{0}^{1} x\left|x-\frac{1}{2}\right| d x\) \(=-\int_{0}^{1 / 2} x\left(x-\frac{1}{2}\right) d x+\int_{1 / 2}^{1} x\left(x-\frac{1}{2}\right) d x\) \(=\int_{0}^{1 / 2}\left(\frac{x}{2}-x^{2}\right) d x+\int_{1 / 2}^{1}\left(x^{2}-\frac{x}{2}\right) d x\) \(=\left[\frac{x^{2}}{4}-\frac{x^{3}}{3}\right]_{0}^{1 / 2}+\left[\frac{x^{3}}{3}-\frac{x^{2}}{4}\right]_{1 / 2}^{1}\) \(\left.=\left(\frac{1}{16}-\frac{1}{24}\right)+\left[\left(\frac{1}{3}-\frac{1}{4}\right)-\frac{1}{24}+\frac{1}{16}\right)\right]\) \(=\left(\frac{6-4}{96}\right)+\left(\frac{32-24-4+6}{96}\right)=\frac{12}{96}=\frac{1}{8}\)
