Chapter 5: Integrals - Section 5.4 - The Fundamental Theorem of Calculus - Exercises 5.4 - Page 287: 35

$\frac{1}{2}(e^1-1)$

Given: $\int_{0}^{1}xe^{x^2}dx ->eqn 1$ Now ,we solve this by using substitution method we take, $x^2=t$ differentiate $wrt . x$ on both sides $=>2xdx=dt$ Now $eqn 1 $ becomes, $\frac{1}{2}\int_{0}^{1}2xe^{x^2}dx=\frac{1}{2}\int_{0}^{1}e^tdt$ $=>\frac{1}{2}[e^t]_{0}^{1}$ $\frac{1}{2}(e^1-e^0)$ $\frac{1}{2}(e^1-1)$