Definite Integration Question 405
Question: If $ \int_0^{1}{x\log ( 1+\frac{x}{2} )}dx=a+b\log \frac{2}{3}, $ then
[SCRA 1986]
Options:
A) $ a=\frac{3}{2},b=\frac{3}{2} $
B) $ a=\frac{3}{4},b=-\frac{3}{4} $
C) $ a=\frac{3}{4},b=\frac{3}{2} $
D) $ a=b $
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Answer:
Correct Answer: C
Solution:
Integrate it by parts taking $ \log ( 1+\frac{x}{2} ) $ as first function $ =[ \log ( 1+\frac{x}{2} )\frac{x^{2}}{2} ]_0^{2}-\int_0^{1}{\frac{1}{1+\frac{x}{2}}\frac{1}{2}\frac{x^{2}}{2}}dx $
$ =\frac{1}{2}\log \frac{3}{2}-\frac{1}{2}\int_0^{1}{\frac{x^{2}}{x+2}dx} $
$ =\frac{1}{2}\log \frac{3}{2}-\frac{1}{2}[ \frac{1}{2}-2+4\log 3-4\log 2 ]=\frac{3}{4}+\frac{3}{2}\log \frac{2}{3} $
On comparing with the given value $ a=\frac{3}{4},b=\frac{3}{2} $ .
BETA
