OpenStudy (anonymous):

Compute the integral of the function over the given region: f(x,y)=(63x+9y)-1, 1≤y≤e, 0≤x≤y f(xy)dA = ?

6 years ago
OpenStudy (anonymous):

$\int\limits_{1}^{e} \int\limits_{0}^{y} (63x+9y-1) dx dy$. Is that all you want?

6 years ago
OpenStudy (anonymous):

yes, I know how to take double integrals but I my problem is how to evaluate it with e and y

6 years ago
OpenStudy (anonymous):

Oh. Okay. Well integrating with respect to x first gives you: $\int\limits_{1}^{e} ((63/2)x^2+9yx-x)|_{0}^{y} dy=\int\limits_{1}^{e} ((63/2)(y^2-0)+9y(y-0)-(y-0)) dy$. If you follow my grouping.

6 years ago
OpenStudy (anonymous):

Then from there you have: $\int\limits_{1}^{e}((63/2)y^2+9y^2-y)dy=(63/6)y^3+3y^3-(1/2)y^2|_{1}^{e}=((63/6)+3)(e^3-1^3)-(1/2)(e^2-1^2)$.

6 years ago
OpenStudy (anonymous):

6 years ago
OpenStudy (anonymous):

Tell me if my grouping made sense. (It throws some people off)

6 years ago
OpenStudy (anonymous):

yes, i think I follow. Hold on let me try to work it on paper. Thanks again!

6 years ago
OpenStudy (anonymous):

No problem :P Try it however you're familiar. I just have a more convenient way (for me at least) of group my terms.

6 years ago
OpenStudy (anonymous):

grouping* v.v

6 years ago
OpenStudy (anonymous):

okay I went through it, I understand what to do with the y now, haha. I ended up with 254.4777 on my calc. My online homework is saying it wrong. Did you come up with a different answer?

6 years ago
OpenStudy (anonymous):

Let me plug it in a calculator h/o

6 years ago
OpenStudy (anonymous):

k thank you!

6 years ago
OpenStudy (anonymous):

Question though. When you typed the problem you had (63x+9y)-1 was that 63x+9y-1 or (63x+9y)^(-1)? HUGE difference lol

6 years ago
OpenStudy (anonymous):

oh no

6 years ago
OpenStudy (anonymous):

haha it was suppose to be ^(-1) !! sorry!!

6 years ago
OpenStudy (anonymous):

Haha, no problem :P This just makes it a usub. So. you see that if you make it: $\int\limits_{1}^{e} \int\limits_{0}^{y}\frac{dx dy}{63x+9y}$. That you can do a u-sub? Let u=63x+9y so du=63 dx or dx=(1/63)du. You see that if you plug in u you get: 1/u? Or, that integrates to ln|u|? (I'm doing this on the side because changing bounds in iterated integrals is a feather). So that becomes: $\frac{1}{63} \int\limits_{1}^{e}(\ln|63x+9y|)|_{0}^{y}dy=\frac{1}{63}\int\limits_{1}^{e}(\ln|63(y)+9y|-\ln|0+9y|)dy$.

6 years ago
OpenStudy (anonymous):

This reduces to: $\frac{1}{63}\int\limits_{1}^{e}(\ln|72y|-\ln|9y|)dy=\frac{1}{63}\int\limits_{1}^{e}\ln(8)dy$. Using the properties of ln's.

6 years ago
OpenStudy (anonymous):

awesome thanks for working with me :), im going go through it on paper again

6 years ago
OpenStudy (anonymous):

No problem :) Then you see though, ln(8) is a constant so you have ln(8)/63 times that integral. Well the integral of dy is just y. So, you have ln(8)/63(e-1). (Hopefully thats right now :P)

6 years ago
OpenStudy (anonymous):

.0567 ! got it! thanks malevolence that was a huge help, I've been stumped on it for the past two days.

6 years ago
OpenStudy (anonymous):

Haha, no problem :D Any question you have for multvar just let me know xP

6 years ago