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Find the areas bounded by the indicated curves.$$y=6-3 x, x=0, y=0, y=3$$

$$A=\frac{9}{2}$$

Calculus 2 / BC

Chapter 26

Applications of Integration

Section 2

Areas by Integration

Harvey Mudd College

University of Michigan - Ann Arbor

Boston College

Lectures

01:11

In mathematics, integratio…

06:55

In grammar, determiners ar…

06:51

Find the areas of the regi…

02:57

09:54

Find the area of the regio…

01:54

01:48

0:00

04:04

03:00

03:23

Use a double integral to c…

03:27

we want to find the area bounded by the curves? Why equal six minus three? X X equals zero, Y equals zero and Y equals three. This question is challenging our understanding of applications of integration in particular how to find the area between curves. So let's not appropriate definitions. The area between two curves A. Is given the vertical components as equal integral to be. Why do you mind if I won the X identically for horizontal components? It's a equal to go see the D extra minus X 24 to identify the appropriate bounds and curbs. To integrate. We're going to sketch this out first, then solve. So sketching this out. We have our area shaded in yellow as the relative region. Why equals 03 and X equals zero are marked. We also have X equal to minus one third. Y marked. To solve what we're going to do is rewrite, X equals two minus one. Rewrite are like 63 X. As I've already having left exactly two minutes one third y and integrate from 0 to 3 from 03 This interval has ended around two y minus y, squared over six, from 0 to 3, or plugging in 4.5.

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