A) \[\frac{4}{3}-\frac{1}{\log 2}\]
B) \[\frac{3}{\log 2}+\frac{4}{3}\]
C) \[\frac{4}{\log 2}-1\]
D) \[\frac{3}{\log 2}-\frac{4}{3}\]
Correct Answer: D
Solution :
Required area =\[\int_{0}^{2}{[{{2}^{x}}-(2x-{{x}^{2}})]\,dx}\] \[=\left[ \frac{{{2}^{x}}}{\log 2}-{{x}^{2}}+\frac{{{x}^{3}}}{3} \right]_{0}^{2}\] \[=\frac{4}{\log 2}-4+\frac{8}{3}-\frac{1}{\log 2}\]\[=\frac{3}{\log 2}-\frac{4}{3}\].
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