# Thread: Change of variable theorem

1. ## Change of variable theorem

I have looked over my notes and the textbook, and the change of variable theorem makes ZERO sense. None. I cannot figure it out. I have tried to ask for help, but this has made me only more confused. Can someone explain IN PLAIN ENGLISH what exactly this thing is saying??

2. ## Re: Change of variable theorem

It is confusing because books suck at explaining it. Here is fast version.

Suppose that $T(u,v)$ is a transformation from $uv$-plane to $xy$-plane given by $T(u,v) = ( g(u,v),h(u,v))$.
Suppose that $D$ is a region in $uv$-plane that gets transformed into region $R$ in $xy$-plane under $T$.
Then,
$$\iint_R f(x,y) = \iint_D f( g(u,v),h(u,v)) ~ |J(T)| ~$$
Where $J(T)$ is the Jacobian,
$$J(T) = \left| \begin{array}{cc} g_u & g_v \\ h_u & h_v \end{array} \right|$$

3. ## Re: Change of variable theorem

Oh it's a double integral. No wonder I was confused. Yeah, the book gave no clear explanation at all.

The example I was given was:

$\int_x^{x+1} \sin \mathrm{e}^t \,\mathrm{d}t = \int_{\mathrm{e}^x}^{\mathrm{e}^{x+1}} \frac{\sin u}{u} \,\mathrm{d}u$

where u = e^t. Can you give me some tips on how to apply what you said to this example?

4. ## Re: Change of variable theorem

Originally Posted by phys251
Oh it's a double integral. No wonder I was confused. Yeah, the book gave no clear explanation at all.

The example I was given was:

$\int_x^{x+1} \sin \mathrm{e}^t \,\mathrm{d}t = \int_{\mathrm{e}^x}^{\mathrm{e}^{x+1}} \frac{\sin u}{u} \,\mathrm{d}u$

where u = e^t. Can you give me some tips on how to apply what you said to this example?
this is pretty straightforward

let $u(t)=e^t$ then $du = e^t~dt \Rightarrow dt=\frac{du}{e^t}=\frac{du}{u}$ so

$\displaystyle{\int_x^{x+1}}\sin(e^t)~dt \Rightarrow \displaystyle{\int_{u(x)}^{u(x+1)}}\sin(u)\frac{du }{u}=$

$\displaystyle{\int_{e^x}^{e^{x+1}}}\frac{\sin(u)} {u}~du$

5. ## Re: Change of variable theorem

Originally Posted by romsek
this is pretty straightforward

let $u(t)=e^t$ then $du = e^t~dt \Rightarrow dt=\frac{du}{e^t}=\frac{du}{u}$ so

$\displaystyle{\int_x^{x+1}}\sin(e^t)~dt \Rightarrow \displaystyle{\int_{u(x)}^{u(x+1)}}\sin(u)\frac{du }{u}=$

$\displaystyle{\int_{e^x}^{e^{x+1}}}\frac{\sin(u)} {u}~du$
That makes perfect sense except for the new limits of integration. How did you know to take u(x+1) and u(x)? (Sorry, like I said, my knowledge here is rusty...)

6. ## Re: Change of variable theorem

that's just what you do. replace the old limits a and b by u(a) and u(b)