2.3 Partielle Integration
Aus Online Mathematik Brückenkurs 2
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Version vom 13:05, 16. Sep. 2008
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Contents:
- Integration by parts.
Learning outcomes:
After this section, you will have learned to:
- Understand the derivation of the formula for integration by parts.
- Solve problems about integration that require integration by parts, followed by a substitution (or vice versa).
Integration by parts
To integrate products, one sometimes can make use of a method known as integration by parts. The method is based on the reverse use of the rules for differentiation of products. If \displaystyle f and \displaystyle g are two differentiable functions then the rule for products gives
- REDIRECT Template:Abgesetzte Formel
Now if one integrates both sides one gets
- REDIRECT Template:Abgesetzte Formel
or after re-ordering
- REDIRECT Template:Abgesetzte Formel
This gives us the formula for integration by parts.
Integration by parts:
- REDIRECT Template:Abgesetzte Formel
This means in practice that one integrates a product of functions by calling one factor \displaystyle f and the other \displaystyle g, and then replaces the integral \displaystyle \,\int f \cdot g\,dx\ , hopefully, by an easier integral \displaystyle \,\int F \cdot g'\,dx\,\mbox{,}\ where \displaystyle F is a primitive function of \displaystyle f and \displaystyle g' is the derivative of \displaystyle g.
It is important to note that the method does not always lead to an integral which is easier than the original. It may also be crucial how one chooses the functions \displaystyle f and \displaystyle g, as the following example shows.
Example 1
Determine the integral \displaystyle \,\int x \cdot \sin x \, dx\,.
If one chooses \displaystyle f=x and \displaystyle g=\sin x one gets \displaystyle F=x^2/2 and \displaystyle g'=\cos x, and the formula for integration by parts gives
- REDIRECT Template:Abgesetzte Formel
The new integral on the right-hand side in this case is not easier than the original integral.
If, instead, one chooses \displaystyle f=\sin x and \displaystyle g=x then \displaystyle F=-\cos x and \displaystyle g'=1, and
- REDIRECT Template:Abgesetzte Formel
Example 2
Determine the integral \displaystyle \ \int x^2 \cdot \ln x \, dx\,.
Put \displaystyle f=x^2 and \displaystyle g=\ln x since differentiation eliminates the logarithm when we carry out an integration by parts: \displaystyle F=x^3/3 and \displaystyle g'=1/x. This gives us that
- REDIRECT Template:Abgesetzte Formel
Example 3
Determine the integral \displaystyle \ \int x^2 e^x \, dx\,.
Put \displaystyle f=e^x and \displaystyle g=x^2, which gives that \displaystyle F=e^x and \displaystyle g'=2x, and an integration by parts gives
- REDIRECT Template:Abgesetzte Formel
This requires further integration by parts to solve the new integral \displaystyle \,\int 2x\,e^x \, dx. We choose in this case \displaystyle f=e^x and \displaystyle g=2x, which gives \displaystyle F=e^x and \displaystyle g'=2
- REDIRECT Template:Abgesetzte Formel
The original integral thus becomes
- REDIRECT Template:Abgesetzte Formel
Example 4
Determine the integral \displaystyle \ \int e^x \cos x \, dx\,.
In a first integration by parts, we have chosen to integrate the factor \displaystyle e^x and differentiate the factor \displaystyle \cos x,
- REDIRECT Template:Abgesetzte Formel
The result of this is that we essentially have replaced the factor \displaystyle \cos x by \displaystyle \sin x in the integral. If we therefore use integration by parts once again (integrate the \displaystyle e^x and differentiate the \displaystyle \sin x) we get
- REDIRECT Template:Abgesetzte Formel
Thus the original integral appears here again. Summarising we have:
- REDIRECT Template:Abgesetzte Formel
and collecting the integrals to one side gives
- REDIRECT Template:Abgesetzte Formel
Although integration by parts in this case did not lead to an easier integral, we arrived at an equation in which the original integral could be ”solved for”. This is not entirely unusual when the integrand is a product of trigonometric functions and / or exponential functions.
Example 5
Determine the integral \displaystyle \ \int_{0}^{1} \frac{2x}{e^x} \, dx\,.
The integral can be rewritten as
- REDIRECT Template:Abgesetzte Formel
Substitute \displaystyle f=e^{-x} and \displaystyle g=2x, and integrate by parts
- REDIRECT Template:Abgesetzte Formel
Example 6
Determine the integral \displaystyle \ \int \ln \sqrt{x} \ dx\,.
We start by performing a variable substitution \displaystyle u=\sqrt{x} which gives \displaystyle du=dx/2\sqrt{x} = dx/2u, that is, \displaystyle dx = 2u\,du\,,
- REDIRECT Template:Abgesetzte Formel
Then we integrate by parts. Put \displaystyle f=2u and \displaystyle g=\ln u, which gives
- REDIRECT Template:Abgesetzte Formel
Note. An alternative approach is to rewrite the initial integrand as \displaystyle \ln\sqrt{x} = \tfrac{1}{2}\ln x and then integrate by parts the product \displaystyle \tfrac{1}{2}\cdot\ln x.
