Definite Integration Question 170
Question: The area of the region enclosed by the curves $ y=x\log x $ and $ y=2x-2x^{2} $ is
Options:
A) $ \frac{5}{12} $
B) $ \frac{7}{12} $
C) 1
D) $ \frac{4}{7} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Curve tracing, $ y=x{\log _{e}}x $ Clearly $ x>0 $ , For $ 0<x<1,x{\log _{e}}x<0, $ and for $ x>1,x{\log _{e}}x>0 $ Also $ x{\log _{e}}x=0 $ or $ x=1 $
Further $ \frac{dy}{dx}=0\Rightarrow 1+{\log _{e}}x=0 $ or $ x=1/e $ , which is point of minima. Required area $ =\int\limits_0^{1}{(2x-2x^{2})dx-\int\limits_0^{1}{x\log xdx}} $
$ =[ x^{2}-\frac{2x^{3}}{3} ]_0^{1}-[ \frac{x^{2}}{2}\log x-\frac{x^{2}}{4} ]_0^{1} $
$ =( 1-\frac{2}{3} )-[ 0-\frac{1}{4}-\frac{1}{2}\underset{x\to 0}{\mathop{\lim }}x^{2}\log x ]=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}. $
BETA
