# Understanding Limits of Integration in Integration-by-Parts

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.

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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

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.

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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 $v = x$.

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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