Lösung 2.3:2b

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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
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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 <math>x^3</math> (so as to reduce its exponent by 1), we need to find a primitive function for <math>e^{x^2}</math>, and how do we do that? If, on the other hand, we integrate <math>x^3</math> and differentiate <math>e^{x^2}</math>, we get
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<math>x^{\text{3}}</math>
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(so as to reduce its exponent by
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<math>\text{1}</math>), we need to find a primitive function for
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<math>e^{x^{2}}</math>, and how do we do that? If, on the other hand, we integrate
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<math>x^{\text{3}}</math>
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and differentiate
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<math>e^{x^{2}}</math>, we get
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{{Displayed math||<math>\begin{align}
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<math>\begin{align}
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\int x^3\cdot e^{x^2}\,dx
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& \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 \\
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&= \frac{x^4}{4}\cdot e^{x^2} - \int\frac{x^4}{4}\cdot e^{x^2}2x\,dx\\[5pt]
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& =\frac{1}{4}x^{4}e^{x^{2}}-\frac{1}{2}\int{x^{5}e^{x^{2}}\,dx} \\
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&= \frac{1}{4}x^{4}e^{x^2} - \frac{1}{2}\int x^5e^{x^2}\,dx
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\end{align}</math>
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\end{align}</math>}}
which just seems to make the integral harder. The solution is instead to substitute
which just seems to make the integral harder. The solution is instead to substitute
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<math>u=x^{2}</math>. If we write the integral as
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<math>u=x^2</math>. If we write the integral as
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<math>\int\limits_{0}^{1}{x^{3}e^{x^{2}}\,dx}=\int\limits_{0}^{1}{x^{2}e^{x^{2}}x\,dx}</math>
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we see that the expression
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<math>''x\,dx''</math>
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can be replaced by
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<math>du</math>
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and the rest of the integrand contains only
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<math>x</math>
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in the form of
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<math>x^{\text{2}}</math>. The substitution gives
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<math>\begin{align}
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{{Displayed math||<math>\int\limits_0^1 x^3e^{x^2}\,dx = \int\limits_0^1 x^2e^{x^2}x\,dx</math>}}
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& \int\limits_{0}^{1}{x^{3}e^{x^{2}}\,dx}=\int\limits_{0}^{1}{x^{2}e^{x^{2}}x\,dx}=\left\{ \begin{matrix}
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u=x^{2} \\
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du=\left( x^{2} \right)^{\prime }\,dx=2x\,dx \\
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\end{matrix} \right\} \\
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& =\int\limits_{0}^{1}{ue^{u}\frac{1}{2}\,du}=\frac{1}{2}\int\limits_{0}^{1}{ue^{u}\,du} \\
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\end{align}</math>
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we see that the expression "<math>x\,dx</math>" can be replaced by <math>du</math>
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and the rest of the integrand contains only <math>x</math> in the form of <math>x^2</math>. The substitution gives
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We can then calculate this integral be partial integration, where we differentiate away the factor
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{{Displayed math||<math>\begin{align}
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<math>u</math>:
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\int\limits_0^1 x^3e^{x^2}\,dx
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&= \int\limits_0^1 x^2e^{x^2}x\,dx\\[5pt]
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&= \left\{\begin{align}
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u &= x^2\\[5pt]
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du &= \bigl(x^2\bigr)'\,dx = 2x\,dx
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\end{align}\right\}\\[5pt]
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&= \int\limits_0^1 ue^u\tfrac{1}{2}\,du\\[5pt]
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&= \frac{1}{2}\int\limits_0^1 ue^u\,du\,\textrm{.}
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\end{align}</math>}}
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We can then calculate this integral by integration by parts, where we differentiate away the factor <math>u</math>,
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<math>\begin{align}
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{{Displayed math||<math>\begin{align}
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& \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} \\
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\frac{1}{2}\int\limits_0^1 ue^u\,du
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& =\frac{1}{2}\left( 1\centerdot e^{1}-0 \right)-\frac{1}{2}\left[ e^{u} \right]_{0}^{1} \\
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&= \frac{1}{2}\Bigl[\ ue^u\ \Bigr]_0^1 - \frac{1}{2}\int\limits_0^1 1\cdot e^u\,du\\[5pt]
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& =\frac{1}{2}e-\frac{1}{2}\left( e^{1}-e^{0} \right) \\
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&= \frac{1}{2}\bigl(1\cdot e^1-0\bigr) - \frac{1}{2}\Bigl[\ e^u\ \Bigr]_0^1\\[5pt]
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& =\frac{1}{2}e-\frac{1}{2}e+\frac{1}{2}=\frac{1}{2} \\
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&= \frac{1}{2}e - \frac{1}{2}\bigl(e^1-e^0\bigr)\\[5pt]
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\end{align}</math>
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&= \frac{1}{2}e - \frac{1}{2}e + \frac{1}{2}\\[5pt]
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&= \frac{1}{2}\,\textrm{.}
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\end{align}</math>}}

Version vom 08:54, 29. Okt. 2008

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}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.3:2b“