Question

Understanding double integrals with u substitution and integral by parts.

I have the double integral from 0 to 9 and from 0 to 1/9x with the equation being e^(x^2) dy dx. I'm not sure how to properly integrate this.

Answer 1

\[\Large\bf \int\limits_{x=0}^9\quad\int\limits_{y=0}^{\frac{1}{9}x}\; e^{x^2}\;dy\;dx\]So we need to integrate this? Hmm let's see...

Answer 2

Yes, I didn't know how to add the equation nicely. I know to take the first integral it just becomes. \[\int\limits_{0}^{9}ye ^{x ^{2}}\] from 0 to 1/9x which becomes \[\int\limits_{0}^{9}(1/9) xe ^{x ^{2}}\] But I do not know how to integrate that properly.

Answer 3

Oh ok good you've got the first part correct :o

Answer 4

Rearranging things a little gives us,
\[\Large\bf \frac{1}{9}\int\limits_{x=0}^9\; e^{\color{#DD4747 }{x^2}}(\color{#3399AA }{x\;dx})\]
We want to make the substitution:\[\Large\bf \color{#DD4747 }{x^2=u}\]So what do we get for du?

Answer 5

du is 2x

Answer 6

\[\Large\bf du=2\;\color{#3399AA }{x\;dx}\]
Hmm ok good.
So we want to solve for our \(\Large\bf \color{#3399AA }{x\;dx}\) so we can replace it in the integral.

Answer 7

Solving for \[xdx\]
it becomes \[\frac{ du }{ 2 } = xdx\]

Answer 8

So the equation would be \[\int\limits_{0}^{9}e ^{x ^{2}}\frac{ du }{ 2 }\]

Answer 9

Don't forget to plug u in as well! :O\[\Large\bf \frac{1}{9}\int\limits\limits_{x=0}^9\; e^{\color{#DD4747 }{x^2}}(\color{#3399AA }{x\;dx})\qquad\to\qquad \frac{1}{9}\int\limits\limits_{x=0}^9\; e^{\color{#DD4747 }{u}}\left(\color{#3399AA }{\frac{1}{2}du}\right)\]

Answer 10

Oh right, I forgot, I'm just having problems visualizing how that becomes.\[\int\limits_{0}^{9}\frac{ 1 }{ 2 } e ^{x ^{2}}\] Also, how are you making the equations bigger?

Answer 11

Or in the end I guess it becomes \[\frac{ 1 }{ 18 }\int\limits_{0}^{9}e ^{x ^{2}}\]

Answer 12

You still didn't plug in the u though D:

Answer 13

In the equation editor, at the start you can type any of these to change the size of the text that follows\[\large \text{\large}, \;\Large\text{\Large},\; \LARGE \text{\LARGE},\;\huge\text{\huge},\;\Huge\text{\Huge}\]

Answer 14

So if i plug in the u I get \[\Large \frac{ 1 }{ 18 }\int\limits_{0}^{9}e^udu\] But how does that get rid of the extra x in du?

Answer 15

This is a tough concept to get comfortable with.
You're not replacing dx with du.
You're replacing the x and dx with something involving du.

And we found the derivative to be,\[\Large\bf du=2\;\color{#3399AA }{x\;dx}\]So dividing through by 2 allows us to "solve for" x dx.\[\Large\bf \color{#3399AA }{\frac{1}{2}du=x\;dx}\]

Answer 16

Let's keep in mind that our bounds are still in terms of x,\[\Large \frac{ 1 }{ 18 }\int\limits\limits_{\color{red}{x=0}}^{9}e^udu\]

So we have a couple of options:
1. You can find new boundaries in terms of u.
2. You can integrate, and then undo your substitution afterwards and plug in the x boundaries.

Answer 17

If we do the second option when we take the antiderivative of \[\Large e ^{u} du\] it just becomes \[\Large e ^{u}\] and from there we would just undo it by putting the x's in so we would get \[\Large \left( \frac{ 1 }{ 18 } \right)\left( e ^{9} - e ^{0} \right)\] Correct?

Answer 18

\[\Large \frac{ 1 }{ 18 }\int\limits\limits\limits\limits_{\color{red}{x=0}}^{9}e^udu\quad=\quad \frac{1}{18}\left(e^u\right)_{x=0}^9\quad=\quad \frac{1}{18}\left(e^{x^2}\right)_{x=0}^9\]

\[\Large =\quad \frac{1}{18}\left(e^{9^2}-e^{0^2}\right)\]

Answer 19

Ah right I forgot the x^2 part, that makes a lot more sense now. Thank You!

Answer 20

\c:/ np