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