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Use a graphing utility, where helpful, to find the area of the region enclosed by the curves.$$x=y^{3}-4 y^{2}+3 y, x=y^{2}-y$$

$$\frac{71}{6}$$

Calculus 1 / AB

Calculus 2 / BC

Chapter 6

APPLICATIONS OF THE DEFINITE INTEGRAL IN GEOMETRY, SCIENCE, AND ENGINEERING

Section 1

Area Between Two Curves

Integrals

Integration

Applications of Integration

Area Between Curves

Volume

Arc Length and Surface Area

Missouri State University

Campbell University

Harvey Mudd College

University of Michigan - Ann Arbor

Lectures

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In mathematics, integratio…

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In grammar, determiners ar…

01:42

Use a graphing utility, wh…

02:16

01:54

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02:30

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Graph the curves $y=x^{2}-…

02:13

Sketch the region enclosed…

07:15

02:22

Use a double integral to c…

eso since you can use a graphing calculator, I'm just gonna go ahead and do that, Um, where you can see my red curve and my blue curve, which match up to these and you can tell there's a cubic and quadratic and they're just sideways, So don't let that freak you up. They intersect at three different places. All you really have to do is figure out ah, the furthest ones away from each other where they intersect. So that is why equals zero must get the white coordinates and y equals four. Um, and what that tells you is the bound. So if you pull up your integral, which is under functions under miscellaneous from 0 to 4. And the reason why we can skip this middle part is I'm gonna absolute value this attraction of the functions and what that does is makes both areas positive. So I don't have to. I'm just saving, um, a lot of time. Now, what you do want to make sure you do is make sure you put parentheses in the second one because you have to make sure that you distribute that minus two each and then tell your culture to do in terms of why, um and this is just a decimal approximation. That looks like 56 though, um, so I got a denominator of six of the 66 plus 5 71 6 Uh, so an exact answer would be some people in six for this. But this calculator just gives you the decibel approximation. There's a lot of thought that gave me this answer. Um, if any of your sitting there and saying, Well, why don't you have to put the first one in parentheses? Um, you don't because there's nothing to distribute in front. But if it helps, you can type it in there, and it will not change your answer. Um, so, yeah, look at your outside bounds. It's all you need for one in a girl for area between these two curves.

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