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Thread: Fairly simple integration by parts query

  1. #1
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    Fairly simple integration by parts query

    Hi all!

    So, I have a question regarding integration by parts, and I think it's pretty straight-forward, but it's bugging the hell out of me...

    Supposing I have an integration $\displaystyle \int_{x_1}^{x_2}f(x)g(x)dx$, and integrate by parts, I'd have $\displaystyle \left[f(x)\int g(x) dx\right]_{x_1}^{x_2} - \int_{x_1}^{x_2}\frac{df(x)}{dx}\int g(x)dxdx$...

    But, should there not be a boundary on the $\displaystyle \int g(x) dx$ terms? If so, what would they be? It just seems odd to me, but I'm probably overlooking something.
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  2. #2
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    Re: Fairly simple integration by parts query

    I would say that no limits are required, because what is required is the antiderivative of g. That is, just rewriting your equation for the purposes of annotation:

    $\displaystyle \int_{x_{1}}^{x_{2}}f(x)\,g(x)\,dx=\Bigg[f(x)\underbrace{\int g(x)\,dx}_{\text{a function}}\Bigg]_{x_{1}}^{x_{2}}-\int_{x_{1}}^{x_{2}}\frac{df(x)}{dx}\underbrace{ \int g(x)\,dx}_{\text{a function}}\,dx.$

    I think it helps to remember from where this formula comes: the product rule for derivatives:

    $\displaystyle (fg)'=f'g+fg'\implies$

    $\displaystyle fg=\int f'g\,dx+\int fg'\,dx\implies$

    $\displaystyle \int fg'\,dx=fg-\int f'g\,dx.$

    Now, the second step there, where I integrated, could have been definite integrals. Then it would have looked like this:

    $\displaystyle [fg]_{x_{1}}^{x_{2}}=\int_{x_{1}}^{x_{2}}f'g\,dx+\int_ {x_{1}}^{x_{2}}fg'\,dx.$

    But notice that both terms in both integrands are just functions. Now the way you've expressed integration by parts, you have my $\displaystyle g'$ equals your $\displaystyle g.$ But they're still just functions.

    Does that help?
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