Integrating $\int^2_0 xe^{x^2}dx$

You mean $\frac12 e^4-\frac12$.

Also, one might set

$g(x) = e^{x^2}; \tag 1$

then

$g'(x) = 2xe^{x^2}; \tag 2$

then

$\displaystyle \int_0^2 xe^{x^2} \; dx = \dfrac{1}{2} \int_0^2 g'(x) \; dx = \dfrac{1}{2}(g(2) - g(0))$ $= \dfrac{1}{2}(e^{2^2} - e^0) = \dfrac{1}{2} (e^4 - 1) = \dfrac{1}{2}(e^4 - 1) = \dfrac{1}{2}e^4 - \dfrac{1}{2}. \tag{3}$

If one wants to use indefinite integrals, we write

$\displaystyle \int xe^{x^2} \; dx = \dfrac{1}{2} \int g'(x) \; dx = \dfrac{1}{2}g(x) + C = \dfrac{1}{2}e^{x^2} + C, \tag 4$

and then proceed to take

$g(2) - g(0) = \dfrac{1}{2}e^4 - \dfrac{1}{2}; \tag 5$

the constant of integration $C$ of course has been cancelled out of this expression.