Question
What is the value of $$\int_0^1 {x{e^{{x^2}}}} dx\,?$$
A.
$$\frac{{\left( {e - 1} \right)}}{2}$$
B.
$${e^2} - 1$$
C.
$$2\left( {e - 1} \right)$$
D.
$$e - 1$$
Answer :
$$\frac{{\left( {e - 1} \right)}}{2}$$
Solution :
$$\eqalign{
& {\text{Let }}I = \int_0^1 {x{e^{{x^2}}}} dx \cr
& {\text{Let }}{x^2} = t \cr
& \Rightarrow 2x\,dx = dt \cr
& \Rightarrow x\,dx = \frac{{dt}}{2} \cr
& {\text{when }}x = 0,\,t = 0{\text{ then }}x = 1,\,t = 1 \cr
& \Rightarrow I = \frac{1}{2}\int\limits_0^1 {{e^t}dt} \cr
& \Rightarrow I = \frac{1}{2}\left[ {{e^t}} \right]_0^1 \cr
& \Rightarrow I = \frac{1}{2}\left[ {{e^{{x^2}}}} \right]_0^1 \cr
& \Rightarrow I = \frac{1}{2}\left[ {e - {e^0}} \right] \cr
& \Rightarrow I = \frac{{e - 1}}{2} \cr} $$