Question

Let $f\left(x\right)=max\{\log (1+x^2),1\}$ for $0\le x\le 2$,

then ${\int }_0^{2}f\left(x\right)dx$ equals?

• A. $\log \left(5/e\right)$
• B. $\log \left(3/e\right)$
• C. ${\tan }^{-1}\sqrt{e}$
• D. None of the above

Answer 1

Answer: (D)

None of the above

${\int }_0^{2}f\left(x\right)dx={\int }_0^{2}max\left[\log (1+x^2),1\right]$

$={\int }_0^{\sqrt{e-1}}1dx+{\int }_{\sqrt{e-1}}^2\log (1+x^2)dx$

$={\left[x\right]}_0^{\sqrt{e-1}}+{\left[x\log (x^2+1)\right]}_{\sqrt{e-1}}^2-{\int }_{\sqrt{e-1}}^2\frac{{2x}^2}{x^2+1}dx$

$=\sqrt{e-1}+2\log 5-\sqrt{e-1}-2{\int }_{\sqrt{e-1}}^2(1-\frac{1}{x^2+1})dx$

$=2\log 5-2{[x-{\tan }^{-1}x]}_{\sqrt{e-1}}^2=2\log 5-2[2-{\tan }^{-1}2-\sqrt{e-1}+{\tan }^{-1}\sqrt{e-1}]=2\log 5-2+2\sqrt{e-1}+2{\tan }^{-1}2+{\tan }^{-1}\sqrt{e-1}$