Lösung 2.3:2b
Aus Online Mathematik Brückenkurs 2
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Version vom 10:27, 11. Mär. 2009
We have a product of two factors in the integrand, so an integration by parts 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^3 (so as to reduce its exponent by 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^3 and differentiate \displaystyle e^{x^2}, we get
| \displaystyle \begin{align}
\int x^3\cdot e^{x^2}\,dx &= \frac{x^4}{4}\cdot e^{x^2} - \int\frac{x^4}{4}\cdot e^{x^2}2x\,dx\\[5pt] &= \frac{1}{4}x^{4}e^{x^2} - \frac{1}{2}\int x^5e^{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^3e^{x^2}\,dx = \int\limits_0^1 x^2e^{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^2. The substitution gives
| \displaystyle \begin{align}
\int\limits_0^1 x^3e^{x^2}\,dx &= \int\limits_0^1 x^2e^{x^2}x\,dx\\[5pt] &= \left\{\begin{align} u &= x^2\\[5pt] du &= \bigl(x^2\bigr)'\,dx = 2x\,dx \end{align}\right\}\\[5pt] &= \int\limits_0^1 ue^u\tfrac{1}{2}\,du\\[5pt] &= \frac{1}{2}\int\limits_0^1 ue^u\,du\,\textrm{.} \end{align} |
We can then calculate this integral by integration by parts, where we differentiate away the factor \displaystyle u,
| \displaystyle \begin{align}
\frac{1}{2}\int\limits_0^1 ue^u\,du &= \frac{1}{2}\Bigl[\ ue^u\ \Bigr]_0^1 - \frac{1}{2}\int\limits_0^1 1\cdot e^u\,du\\[5pt] &= \frac{1}{2}\bigl(1\cdot e^1-0\bigr) - \frac{1}{2}\Bigl[\ e^u\ \Bigr]_0^1\\[5pt] &= \frac{1}{2}e - \frac{1}{2}\bigl(e^1-e^0\bigr)\\[5pt] &= \frac{1}{2}e - \frac{1}{2}e + \frac{1}{2}\\[5pt] &= \frac{1}{2}\,\textrm{.} \end{align} |
