   Chapter 5.1, Problem 28E

Chapter
Section
Textbook Problem

# Sketch the region enclosed by the given curves and find its area. y = 1 4 x 2 ,   y = 2 x 2 ,   x + y = 3 ,   x ≥ 0

To determine

To:

Sketch the region and find the enclosed area.

Explanation

1) Concept:

Formula:

The area A of the region bounded by the curves   y=f(x), y=g(x) and the lines x=a and x=b  is

A= abfx-gxdx

fx-gx=fx-gx when fxg(x)gx-fx when gxf(x)

2) Given:

y=14x2,  y=2x2 and   x+y=3,    x0

3) Calculation:

Write   x+y=3 as y in terms of x.

y=3-x

The point of intersection occurs when both the equation are equal to each other.

So intersection of    y=14x2 and   y=2x2 is given by

14x2=2x2

8x2-x2=0

7x2=0

that is at x=0

Intersection of   y=14x2 and   y=3-x

14x2=3-x

x2+4x-12=0

x-2x+6=0

that is at x=2 and   x=-6

Intersection point of   y=2x2 and   y=3-x

2x2=3-x

2x2+x-3=0

2x+3x-1=0

that is,  x=1 and   x=-32

Thus, the points of intersection are at x=0, x=1 and   x=2 as   x0. The region is sketched in the following figure.

Here,   A=A1+A2

where A1 is the area between the curves from  0x1

and A2 is the area between the curves from  1x2

Here,2x214x2 when   0x1. Therefore, let’s assume that

fx=2x2

gx=14x2

Therefore, the required area is

A1=012x2-14x2dx

A1=012x2-14x2 dx

A1=0174x2 dx

Compute the integral using the standard integration rule

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