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Lösung 2.3:2b

Aus Online Mathematik Brückenkurs 2

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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 x3 (so as to reduce its exponent by 1), we need to find a primitive function for ex2, and how do we do that? If, on the other hand, we integrate x3 and differentiate ex2, we get

x3ex2dx=4x4ex2−4x4ex22xdx=41x4ex2−21x5ex2dx

which just seems to make the integral harder. The solution is instead to substitute u=x2. If we write the integral as

10x3ex2dx=10x2ex2xdx 

we see that the expression "xdx" can be replaced by du and the rest of the integrand contains only x in the form of x2. The substitution gives

10x3ex2dx=10x2ex2xdx=udu=x2=x2dx=2xdx=10ueu21du=2110ueudu.

We can then calculate this integral by integration by parts, where we differentiate away the factor u,

2110ueudu=21 ueu 10−21101eudu=211e1−0−21 eu 10=21e−21e1−e0=21e−21e+21=21.
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.3:2b“