question_answer

The area of the region enclosed by the curves \[y=x\,\,\log \,\,x\] and \[y=2x-2{{x}^{2}}\] is  

A) \[\frac{5}{12}\]

B) \[\frac{7}{12}\]

C) 1

D) \[\frac{4}{7}\]

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 \[=\int\limits_{0}^{1}{(2x-2{{x}^{2}})dx-\int\limits_{0}^{1}{x\log x\,dx}}\] \[=\left[ {{x}^{2}}-\frac{2{{x}^{3}}}{3} \right]_{0}^{1}-\left[ \frac{{{x}^{2}}}{2}\log x-\frac{{{x}^{2}}}{4} \right]_{0}^{1}\] \[=\left( 1-\frac{2}{3} \right)-\left[ 0-\frac{1}{4}-\frac{1}{2}\underset{x\to 0}{\mathop{\lim }}\,{{x}^{2}}\log x \right]=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}.\]