Question

${\int }_0^1(1+e^{-x^2})dx$ is equal to

• A. $-1$
• B. $2$
• C. $1+e^{-1}$
• D. None of these

Answer 1

The correct option is D None of these

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

$I={\int }_0^1\left(1dx\right)+{\int }_0^1{e}^{-x^2}dx$
Now multiply and divide Second term by $\frac{\sqrt{\pi }}{2}$

$I=|x{|}_0^1+\frac{\sqrt{\pi }}{2}{\int }_0^1\frac{2e^{-x^2}}{\sqrt{\pi }}dx$

Second one is special integral(Gaussian error function) and it's value is $erf\left(x\right)$

$\therefore I=1+\frac{\sqrt{\pi }}{2}|erf\left(x\right){|}_0^1=1+\frac{\sqrt{\pi }}{2}erf\left(1\right)$

Hence, $\left(D\right)$