Aus Online Mathematik Brückenkurs 2

We have a product of two factors in the integrand, so a partial integration does not seem unreasonable. There is nevertheless a problem as regards which factor should be differentiated and which should be integrated. If we choose to differentiate \displaystyle x^{\text{3}} (so as to reduce its exponent by \displaystyle \text{1}), we need to find a primitive function for \displaystyle e^{x^{2}}, and how do we do that? If, on the other hand, we integrate \displaystyle x^{\text{3}} and differentiate \displaystyle e^{x^{2}}, we get

\displaystyle \begin{align} & \int{x^{3}e^{x^{2}}\,dx=\frac{x^{4}}{4}}\centerdot e^{x^{2}}-\int{\frac{x^{4}}{4}}\centerdot e^{x^{2}}2x\,dx \\ & =\frac{1}{4}x^{4}e^{x^{2}}-\frac{1}{2}\int{x^{5}e^{x^{2}}\,dx} \\ \end{align}

which just seems to make the integral harder. The solution is instead to substitute \displaystyle u=x^{2}. If we write the integral as

\displaystyle \int\limits_{0}^{1}{x^{3}e^{x^{2}}\,dx}=\int\limits_{0}^{1}{x^{2}e^{x^{2}}x\,dx}

we see that the expression \displaystyle ''x\,dx'' can be replaced by \displaystyle du and the rest of the integrand contains only \displaystyle x in the form of \displaystyle x^{\text{2}}. The substitution gives

\displaystyle \begin{align} & \int\limits_{0}^{1}{x^{3}e^{x^{2}}\,dx}=\int\limits_{0}^{1}{x^{2}e^{x^{2}}x\,dx}=\left\{ \begin{matrix} u=x^{2} \\ du=\left( x^{2} \right)^{\prime }\,dx=2x\,dx \\ \end{matrix} \right\} \\ & =\int\limits_{0}^{1}{ue^{u}\frac{1}{2}\,du}=\frac{1}{2}\int\limits_{0}^{1}{ue^{u}\,du} \\ \end{align}

We can then calculate this integral be partial integration, where we differentiate away the factor \displaystyle u:

\displaystyle \begin{align} & \frac{1}{2}\int\limits_{0}^{1}{ue^{u}\,du}=\frac{1}{2}\left[ ue^{u} \right]_{0}^{1}-\frac{1}{2}\int\limits_{0}^{1}{1\centerdot e^{u}\,du} \\ & =\frac{1}{2}\left( 1\centerdot e^{1}-0 \right)-\frac{1}{2}\left[ e^{u} \right]_{0}^{1} \\ & =\frac{1}{2}e-\frac{1}{2}\left( e^{1}-e^{0} \right) \\ & =\frac{1}{2}e-\frac{1}{2}e+\frac{1}{2}=\frac{1}{2} \\ \end{align}