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Describe a region for which the area is found by evaluating the integral $\int_{1}^{2}\left(2 x^{2}-x^{3}\right) d x$

the area bounded by $x=1, y=0,$ and $y=2 x^{2}-x^{3}$.

Calculus 2 / BC

Chapter 26

Applications of Integration

Section 2

Areas by Integration

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we want to determine the value of this integral, um, the integral from minus 1 to 1 of the absolute value of three to the ex miners to to the ex. So let's do a quick sketch of our functions inside. Three to the X. Looks like that. Um, okay, three to the ex and then two to the ex Arises a bit most deeply. I mean, a little bit less deeply. So two to the ex. An instance. We're going from minus 1 to 1. Let's draw daughter lines one and one. We find that the the integral is going to give us the total area enclosed. Ah, here in red. So we're going to have to break up this integral at this point over here. So we see that these two functions cross at the origin and only at the origin Essence three to the X equals two to the ex. I tells us that while taking the natural lager them we have X long three equals x salon, too, which tells us that X equals zero. So these two functions only cross at the origin so we can break up this integral at X equals zero. So here. We're integrating from negative 1 to 0 of our top function, which is to to the ex minus three to the X D x and then to the right of the origin. We have, um, integral from 0 to 1 of three to the X minus two to the ex DX. And this is what we want to evaluate. Um, this. So let's do that on the next page. Uh, we have equals integral zero to know it was minus 1 to 0 of two to the X minus three to the ex DX plus integral 0 to 1 of three to the X minus two to the x d X. Let's just carry out these anti derivatives. The anti derivative of two to the eggs is to to the ex over lawn of two and then the same thing. Here it's three to the X over long three. This is going from minus 1 to 0. And then here we have three to the X over long, three minus two to the X over lung, too, going from 0 to 1 so plugging in our values. So when we plug in X equals zero here we get one over lunch too. Minus one over long. Three over this three years. Okay. And then we subtract plugging minus one. So that becomes too to the minus one. So it's, like, too long to hear in the denominator. And then similarly, we have three long three here in the denominator, plus plugging in one. We get three over long three, um, and then plugging in one again. Oh, here we have minus two over lunch, too. Now, plugging in the zero. Subtract three to the zero is one over long. Three minus won over London two. Okay, so this is what we want to simplify. So let's do it. After we simplify everything we get that the value equals for four. Okay? It's not writing. Hold on. Let's open up the next page for over three. Lung of three, minus one over to loan of two. So this is the area of the enclosed region

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