Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

My understanding of integration-by-parts is a little shaky. In particular, I'm not totally certain that I understand how to properly calculate the limits of integration.

For example, the formula I have is:

$\int_{v_1}^{v_2}{u dv} = (u_2 v_2 - u_1 v_1) - \int_{u_1}^{u_2}{v du}$

I'd like to see how to calculate $u_1$ and $u_2$, preferably in a complete example (that solves a definite integral.) I'm really interested in an example where the limits of integration change; i.e. $u_1$ and $u_2$ are different than $v_1$ and $v_2$, if possible.

share|improve this question
    
The limits of integration in both sides, although different in terms of $u(x),v(x)$, when expressed in terms of the same variable $x$ of functions $u(x),v(x)$ are equal. –  Américo Tavares Dec 1 '10 at 18:40
    
@Américo Tavares: Thanks. This was one of my main concerns, but I wasn't sure exactly which cases this was valid for. –  Matt Groff Dec 1 '10 at 21:42

2 Answers 2

up vote 12 down vote accepted

A more precise notation is this one

$$\int_{x_{1}}^{x_{2}}u(x)v^{\prime }(x)dx=\left( u(x_{2})v(x_{2})-u(x_{1})v(x_{2})\right) -\int_{x_{1}}^{x_{2}}u^{\prime }(x)v(x)dx$$

which is derived from the derivative rule for the product

$$(u(x)v(x))^{\prime }=u^{\prime }(x)v(x)+u(x)v^{\prime }(x)$$

or

$$u(x)v^{\prime }(x)=(u(x)v(x))^{\prime }-u^{\prime }(x)v(x).$$

So

$$\begin{eqnarray*} \int_{x_{1}}^{x_{2}}u(x)v^{\prime }(x)dx &=&\int_{x_{1}}^{x_{2}}(u(x)v(x))^{\prime }dx-\int_{x_{1}}^{x_{2}}u^{\prime }(x)v(x)dx \\ &=&\left. (u(x)v(x))\right\vert _{x=x_{1}}^{x=x_{2}}-\int_{x_{1}}^{x_{2}}u(x)v(x)dx \\ &=&\left( u(x_{2})v(x_{2})-u(x_{1})v(x_{2})\right) -\int_{x_{1}}^{x_{2}}u^{\prime }(x)v(x)dx. \end{eqnarray*}.$$

If you write $dv=v^{\prime }(x)dx$ and $du=u^{\prime }(x)dx$, you get your formula but with $u,v$ as a function of $x$

$$\int_{v_{1}(x)}^{v_{2}(x)}u(x)dv=\left( u(x_{2})v(x_{2})-u(x_{1})v(x_{2})\right) -\int_{u_{1}(x)}^{u_{2}(x)}v(x)du$$

Example: Assume you want to evaluate $\int_{x_{1}}^{x_{2}}\log xdx=\int_{x_{1}}^{x_{2}}1\cdot \log xdx$. You can choose $v^{\prime }(x)=1$ and $u(x)=\log x$. Then $v(x)=x$ (omitting the constant of integration) and $u^{\prime }(x)=\frac{1}{x}$. Hence

$$\begin{eqnarray*} \int_{x_{1}}^{x_{2}}\log xdx &=&\int_{x_{1}}^{x_{2}}1\cdot \log xdx \\ &=&\left( \log x_{2}\cdot x_{2}-\log x_{1}\cdot x_{1}\right) -\int_{x_{1}}^{x_{2}}\frac{1}{x}\cdot xdx \\ &=&\left( \log x_{2}\cdot x_{2}-\log x_{1}\cdot x_{1}\right) -\int_{x_{1}}^{x_{2}}dx \\ &=&\left( \log x_{2}\cdot x_{2}-\log x_{1}\cdot x_{1}\right) -\left( x_{2}-x_{1}\right) \end{eqnarray*}$$


The same example with your formula:

$$u=\log x,v=x,dv=dx,v=x,du=\frac{1}{x}dx$$

$$u_{2}=\log x_{2},u_{1}=\log x_{1},v_{2}=x_{2},v_{1}=x_{1}$$

$$\begin{eqnarray*} \int_{v_{1}}^{v_{2}}udv &=&\left( u_{2}v_{2}-u_{1}v_{2}\right) -\int_{u_{1}}^{u_{2}}vdu \\ \int_{x_{1}}^{x_{2}}\log xdx &=&\left( \log x_{2}\cdot x_{2}-\log x_{1}\cdot x_{1}\right) -\int_{\log x_{1}}^{\log x_{2}}xdu \\ &=&\left( \log x_{2}\cdot x_{2}-\log x_{1}\cdot x_{1}\right) -\int_{x_{1}}^{x_{2}}x\cdot \frac{1}{x}dx \\ &=&\left( \log x_{2}\cdot x_{2}-\log x_{1}\cdot x_{1}\right) -\left( x_{2}-x_{1}\right). \end{eqnarray*}$$

Note: The limits of integration, although different in terms of $u(x),v(x)$, when expressed in terms of the same variable $x$ of functions $u(x),v(x)$ are the same in both sides.

For a strategy on how to chose the $u$ and $v$ terms see this question.

share|improve this answer

Okay.

$$\int_1^2 \ln x \, dx = [x \ln x]_{x = 1}^2 - \int_1^2 1 \, dx = 2 \ln 2 - 1$$

A prototype example. Where $u = \ln x$ and $v = x$.

share|improve this answer
    
I wasn't very clear. I'm going to change my question. I'd like to see an example where the limits of integration change. –  Matt Groff Dec 1 '10 at 17:41
    
@Matt Groff: Okay, you also want some substitution to compute the second integral? –  Jonas Teuwen Dec 1 '10 at 18:06
    
I'm interested in the case when the limits change. I'm attempting to avoid the situation when the limits change, but I wasn't totally certain when this happens. I also would like to understand the change of limits anyways, in case I use it someday. –  Matt Groff Dec 1 '10 at 23:36

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.