Question

Let $Ι_{1}=\displaystyle \int _{0}^{1} e^{x^{2}} d x$ and $Ι_{2}=\displaystyle \int _{0}^{1} 2 x^{2} e^{x^{2}} d x$ , then the value of $Ι_{1}+Ι_{2}$ is equal to

• A. $1$
• B. $2$
• C. $e$
• D. $e^{2}$

Answer 1

Answer: (C)

$Ι_{2}=\displaystyle \int _{0}^{1} 2 x^{2} e^{x^{2}} d x$
$=\displaystyle \int _{0}^{1}\left(2 x e^{x^{2}}\right)xdx$
Using integration by parts,
$Ι_{2}=\left[x e^{x^{2}}\right]_{0}^{1}-\displaystyle \int _{0}^{1} e^{x^{2}} d x$
$Ι_{2}+Ι_{1}=e$