Understanding Limits of Integration in Integration - by - Parts

My understanding of integration - by - components is a little unsteady. Specifically, I'm not entirely particular that I recognize just how to effectively compute the restrictions of integration.

As an 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 would certainly such as to see just how to compute $u_1$ and also $u_2$, ideally in a full instance (that addresses a precise indispensable.) I'm actually curious about an instance where the restrictions of integration adjustment ; i.e. $u_1$ and also $u_2$ are various than $v_1$ and also $v_2$, when possible.

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2019-12-02 02:51:06
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Answers: 2

Okay.

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

A model instance. Where $u = \ln x$ and also $v = x$.

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2019-12-03 04:24:13
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An extra specific symbols is this set

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

which is stemmed from the acquired regulation for the item

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

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 also $du=u^{\prime }(x)dx$, you get your formula yet 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 intend to review $\int _ x _ 1 ^ x _ 2 \log xdx = \int _ x _ 1 ^ x _ 2 1\cdot \log xdx$. You can choose $v ^ \prime (x) = 1$ and also $u(x)=\log x$. After that $v(x)=x$ (leaving out the constant of integration) and also $u^{\prime }(x)=\frac{1}{x}$. Therefore

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


The very same instance 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 restrictions of integration, although various in regards to $u(x),v(x)$, when shared in regards to the very same variable $x$ of features $u(x),v(x)$ are the very same in both sides.

For an approach on just how to picked the $u$ and also $v$ terms see this question.

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2019-12-03 04:23:29
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