Processing Math: Done
To print higher-resolution math symbols, click the
Hi-Res Fonts for Printing button on the jsMath control panel.
No jsMath TeX fonts found -- using image fonts instead.
These may be slow and might not print well.
Use the jsMath control panel to get additional information.
jsMath Control PanelHide this Message

jsMath

Lösung 2.3:2b

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K
K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
Zeile 1: Zeile 1:
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
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
-
{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\int x^3\cdot e^{x^2}\,dx
\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{x^4}{4}\cdot e^{x^2} - \int\frac{x^4}{4}\cdot e^{x^2}2x\,dx\\[5pt]
Zeile 10: Zeile 10:
<math>u=x^2</math>. If we write the integral as
<math>u=x^2</math>. If we write the integral as
-
{{Displayed math||<math>\int\limits_0^1 x^3e^{x^2}\,dx = \int\limits_0^1 x^2e^{x^2}x\,dx</math>}}
+
{{Abgesetzte Formel||<math>\int\limits_0^1 x^3e^{x^2}\,dx = \int\limits_0^1 x^2e^{x^2}x\,dx</math>}}
we see that the expression "<math>x\,dx</math>" can be replaced by <math>du</math>
we see that the expression "<math>x\,dx</math>" can be replaced by <math>du</math>
and the rest of the integrand contains only <math>x</math> in the form of <math>x^2</math>. The substitution gives
and the rest of the integrand contains only <math>x</math> in the form of <math>x^2</math>. The substitution gives
-
{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\int\limits_0^1 x^3e^{x^2}\,dx
\int\limits_0^1 x^3e^{x^2}\,dx
&= \int\limits_0^1 x^2e^{x^2}x\,dx\\[5pt]
&= \int\limits_0^1 x^2e^{x^2}x\,dx\\[5pt]
Zeile 28: Zeile 28:
We can then calculate this integral by integration by parts, where we differentiate away the factor <math>u</math>,
We can then calculate this integral by integration by parts, where we differentiate away the factor <math>u</math>,
-
{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\frac{1}{2}\int\limits_0^1 ue^u\,du
\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[\ ue^u\ \Bigr]_0^1 - \frac{1}{2}\int\limits_0^1 1\cdot e^u\,du\\[5pt]

Version vom 13:04, 10. 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 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“