Let $I=\int_0^1 e^{x^2} \cdot x^3 d x=\int_0^1 e^{x^2} \cdot x^2 \cdot x d x$ Put $x^2=t \quad \therefore 2 x d x=d t$ $\therefore x d x=\frac{d t}{2}$ When $x=0, t=0$ When $x=1, t=1$ $ \begin{aligned} \therefore I & =\int_0^1 e^t \cdot t \cdot \frac{d t}{2}=\frac{1}{2} \int_0^1 t e^t d t \\ & =\frac{1}{2}\left\{\left[t \int e^t d t\right]_0^1-\int_0^1\left[\frac{d}{d t}(t) \int e^t d t\right] d t\right\} \\ & =\frac{1}{2}\left[t e^t\right]_0^1-\frac{1}{2} \int_0^1 1 \cdot e^t d t \\ & =\frac{1}{2}(e-0)-\frac{1}{2}\left[e^t\right]_0^1 \\ & =\frac{e}{2}-\frac{1}{2}(e-1) \\ & =\frac{e}{2}-\frac{e}{2}+\frac{1}{2}=\frac{1}{2} \end{aligned} $
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