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I haven't studied multiple integrals yet, but when I look at the integration by parts formula, I see that there is an integral of an integral. If the integration being done has limits, I'm not sure how they're managed. Here is what I have:

So the by parts formula is $\int f(x)g'(x)dx = f(x)g(x)-\int f'(x)g(x)dx$

If I have $g'(x)$, then I need $g(x)$.

My question is this: if I'm solving with limits, do I apply limits to $\int g'(x)dx$ also?

In other words, do I do this

$\int_a^b xe^{6x}dx = x \cdot |\frac{1}{6}e^{6x}|_a^b - \int_a^b1\cdot \frac{1}{6}e^{6x}dx$

or this

$\int_a^b xe^{6x}dx = x \cdot |\frac{1}{6}e^{6x}|_a^b - \int_a^b1\cdot |\frac{1}{6}e^{6x}|_a^bdx$


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The correct formula is $[x(1/6)e^{6x}]|_a^b -\int_a^b e^{6x}dx$. In the IPB formula $V=g(x)$ is a specific antiderivative, not a definite integral. The general rule would be $\int_a^b f(x)g'(x)\, dx=[f(x)g(x) ]|_a^b - \int _a^b g(x)f'(x) \,dx$. – David Mitra Feb 8 '12 at 21:12
This is not a double integral. An example of a double integral is $\int_{a}^{b}\int_{c}^{d}f(x,y)dxdy$. – Américo Tavares Feb 8 '12 at 21:21
@AméricoTavares Ah, ok. Thanks. – Korgan Rivera Feb 8 '12 at 21:26
@DavidMitra why is x included in the limits for the integral of $e^{6x}$? Why isn't it $x[(1/6)e^{6x}]|_a^b$?...never mind I figured it out :) Thanks! – Korgan Rivera Feb 8 '12 at 21:28
@Korgan Because the IPB formula says an antiderivative of $fg'$ is $fg-\int gf'$. So, to evaluate a definite integral $\int_a^b f(x)g'(x)\,dx$: You first find an antiderivative of the integrand. One is, by the IPB formula, $F (x) =f(x)g(x)-\int g(x)f'(x)\, dx$. Then, you plug in the limits and take the difference. The result of this is equivalent to what the formula I have above gives. – David Mitra Feb 8 '12 at 21:33
up vote 1 down vote accepted

It goes like this:

$\int_a^b xe^{6x}dx = (x \cdot \frac{1}{6}e^{6x})|_a^b - \int_a^b(1\cdot \frac{1}{6}e^{6x})dx$

$f(x) g(x)$ is evaluated from $a$ to $b$ and so is the integral $\int^b_a [f'(x) g(x)] dx$

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