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Use the region $R$ with the indicated boundaries to evaluate each double integral.

$$

\iint_{R}(5 x+8 y) d y d x ; \quad 1 \leq x \leq 3,0 \leq y \leq x-1

$$

34

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Harvey Mudd College

University of Nottingham

Idaho State University

Boston College

Okay, so we're giving this double Integral, um, of region are five x plus y de y de axe, and we want to enrage over a somewhat odd region. It's basically when X is between one and three on when wise between zero and X minus one and drawing this out, this graph is gonna kind of look like this. So we have This is the y equals X minus one line and then zero y equals zero and then exits between one and three. So is this little triangular reason here? Uh, basically, we need to find the double integral over this writing or this triangular region so ready in the limits is a little tricky, but it should be. I'll explain the thought process, make it pretty simple. So are we always evaluate intervals from the inside to the outside. So we think about how each variable ranges and we think about how y ranges first. Now, why ranges from wizards who X minus one? Basically, we enter from the bottom at X equals zero all the way to X minus one, and we have to keep our limit kind of variable like as a variable, because our endpoint always changes. And it depends on where you're at along this line. So we keep it as X minus one and then our X ranges from 13 from one, 23 obviously. Ah, yeah. Wait tonight. Why ranges from here to here? Yeah, I'm right. Okay. So evaluating this inter girl, um, we want to integrate the inside first. So we have five X plus eight. Y de y um, Okay. Yeah. Sorry. Um, let me. Okay, So since we're in agreeance but to whyfors, um, you think about other every other Babel as a constant. So we have basically excess constant. So we have five x. Why? Plus eight y squared over two from zero to X minus one, and then some. Find this a little bit. We have five x y plus four y squared from zero to X minus one. Um, and then plugging this in since we're in your inspector, Why we plug into Why? So we have five x times X minus one, and then plus four times x minus 1 20 squared. And then we were surprised by what we get when we plug in zero and should just be served because we have wise and both of them. And the multiplication by zero is always zero. So now just we just want some. Fight us a little bit So five X squared minus five X plus four times X squared minus two X plus one A sequel to five X squared minus five X plus four X squared minus eight X plus four Sequel to nine X squared, minus 13 x plus before okay, and now we plug this back into our outer integral. So in a row from one of three of nine x squared minus 13 x plus four d X Um, OK, and then valuing this, we use the reverse spiral. So next the third over three plus o minus 13 x squared over two plus four x from 1 to 3. Some five this a little bit uh, get three x the third minus 13 x squared over two plus four x 123 and just plugging in. We have three times Street to the third, which is just three to the fourth and three to the fourth is 81 minus 13 times nine over to plus 12 and then minus three miles, 13/2 plus four on. If you just work it out, it should just be 34. And that is our final answer. I think. Um Yep. This should be our final answer.

Rutgers, The State University of New Jersey