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Use a graphing utility, where helpful, to find the area of the region enclosed by the curves.$$x=y^{3}-y, x=0$$

$$\frac{1}{2}$$

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

Baylor University

University of Michigan - Ann Arbor

Lectures

01:11

In mathematics, integratio…

06:55

In grammar, determiners ar…

01:54

Use a graphing utility, wh…

01:42

02:00

02:16

02:06

02:30

00:36

Sketch the region enclosed…

04:00

03:23

Use a double integral to c…

01:57

eso won and graft this equation already be careful. Sonata functions You don't wanna say that. And then X equals zero is just the y axis. You get a visual there, and what's important is you want to see that the bounds that where they cross are from negative one, they do cross at zero. But I'm not gonna do that with my integral show. You a way around that, um, and positive one. And the reason why I'm going to ignore that is I just copy this equation real quick if I were to create a new equation, but right as the absolute value of that function, you can actually type into your calculator so the areas will not cancel each other out. Eso then all you'd have to do is find your integral key, which mines under the functions miscellaneous and egg roll from negative one to one of the absolute value of that function. And then you after I d y, um, and you get the exact answer of only five. Just so you know, there's other ways of doing this. If I hide that function and go back to this, um, some of you might be saying there. What if you go from negative 120 You can absolutely no 1 to 0. But think about. Okay, well, we want the area between the Kurds and areas always positive. So then double that. That's why those areas you do not want to cancel out family was thinking that, um ah, and you can actually do the integral of this. It's 1/4 y to the fourth minus one half life squared and plug in one and negative one and see that you'll get possible in half. I feel like I've over explained this problem to this point, so hopefully you're happy with what I've said so far.

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