Definite integral of exponential function $se^{s^{2}}$

Question

I'm trying to figure out how to solve this but haven't come up with anything yet:

$$\int\limits_0^t{se^{s^{2}}}ds$$

The solution I wrote down is:

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

Can anyone tell me how I can get to this solution?

Thanks a lot.

Answer 1

$\int\limits_{0}^{t} se^{s^{2}}ds=\frac{1}{2}\int\limits_{0}^{t} 2s e^{s^{2}} ds=\frac{1}{2}[e^{s^{2}}]_{0}^{t}=\frac{1}{2}(e^{t^{2}}-1)$

Answer 2

To solve the integral you can use change of variables :\

\begin{eqnarray} \int_0^t s e^{s^2}ds &{}& \nonumber \\ \text{consider the chage: } \nonumber \\ x = s^2 &\to& dx = 2sds,\text{ then:} \nonumber\\ \int_0^t s e^{s^2}ds &\to& \int_0^{t^2} \frac{1}{2} e^{x}dx \nonumber\\ &=& \frac{1}{2} \Big( e^{t^2} -1 \Big)_{\square} \end{eqnarray}