question_answer

The area of the region bounded by the straight lines \[x=0\]and \[x=2,\] and the curves \[y={{2}^{x}}\] and \[y=2x-{{x}^{2}}\] is equal to :

A)  \[\frac{2}{\log 2}-\frac{4}{3}\]

B)  \[\frac{3}{\log 2}-\frac{4}{3}\]

C)  \[\frac{1}{\log 2}-\frac{4}{3}\]

D)  \[\frac{4}{\log 2}-\frac{3}{2}\]

Correct Answer: B

Solution :

Required area \[=\int_{0}^{2}{[{{2}^{x}}-(2x-{{x}^{2}})]\,dx}\] \[=\int_{0}^{2}{({{2}^{x}}-2x+{{x}^{2}})\,dx}\] \[=\left[ \frac{{{2}^{x}}}{\log 2}-{{x}^{2}}+\frac{{{x}^{3}}}{3} \right]_{0}^{2}\] \[=\frac{4}{\log 2}-4+\frac{8}{3}-\frac{1}{\log 2}\] \[=\left( \frac{3}{\log 2}-\frac{4}{3} \right)\text{sq}\,\text{unit}\]