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Math Help - Change of variable theorem

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    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??
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    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,
    $$ J(T) = \left| \begin{array}{cc} g_u & g_v \\ h_u & h_v \end{array} \right| $$
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    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?
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    Re: Change of variable theorem

    Quote Originally Posted by phys251 View Post
    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

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

    $\displaystyle{\int_{e^x}^{e^{x+1}}}\frac{\sin(u)} {u}~du$
    Thanks from phys251
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    Re: Change of variable theorem

    Quote Originally Posted by romsek View Post
    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...)
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    Re: Change of variable theorem

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