Question

Compute the area of the figure bounded by straight lines $x=0$, $x=2$ and the curves $y=2^x$ and $y=2x-x^2$.

• A. $\frac{3}{\log 2}-\frac{4}{3}$
• B. $\frac{3}{\log 2}+\frac{4}{3}$
• C. $\frac{4}{\log 3}-\frac{4}{3}$
• D. $\frac{4}{\log 2}+\frac{1}{3}$

Answer 1

The correct option is A $\frac{3}{\log 2}-\frac{4}{3}$

$={\int }_0^2{2}^x-2x+x^{2}dx$

$={[\frac{2^x}{\ln \left(2\right)}+\frac{(x-3)x^2}{3}]}_0^2$

$=[\frac{4}{\ln 2}-\frac{4}{3}]-\left[\frac{1}{\ln 2}\right]$

$=\frac{3}{\log 2}-\frac{4}{3}$