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Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

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Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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Checkpoint 4 Worked-out solution available at LarsonAppliedCalculus.com

Find the area of the region bounded by the graphs of

f ( x ) = x 3 + 2 x 2 3 x  and g ( x ) = x 2 + 3 x

Sketch a graph of the region.

To determine

To calculate: The area of the region bounded by the graphs of f(x)=x3+2x23x and g(x)=x2+3x and then sketch the graph.

Explanation

Given Information:

The two functions f(x)=x3+2x23x and g(x)=x2+3x.

Formula used:

The area of a region bounded by two graphs is calculated using the following formula,

If f and g are continuous on [a,b] and g(x)f(x) for all x in [a,b], then the area of the region bounded by the graphs of f, g, x=a and x=b is given by

A=ab[f(x)g(x)]dx

The power rule is xndx=xn+1n+1+C,n1.

Calculation:

Consider the provided functions,

f(x)=x3+2x23x

And, g(x)=x2+3x

Compute the points of intersection of two graphs by setting the functions equal to each other and solving for x,

f(x)=g(x)x3+2x23x=x2+3xx3+2x2x23x3x=0x3+x26x=0

Factor the above polynomial,

x(x2+x6)=0x(x2+3x2x6)=0x[x(x+3)2(x+3)]=0x(x+3)(x2)=0

That is,

x=0x+3=0x=3

And,

x2=0x=2

Substitute x=3 in the function f(x)=x3+2x23x and compute the first intersection point,

f(3)=(3)3+2×(3)23×(3)=27+18+9=0

Substitute x=0 in the function f(x)=x3+2x23x and compute the second intersection point,

f(0)=03+2×023×0=0+00=0

Substitute x=2 in the function f(x)=x3+2x23x and compute the third intersection point,

f(2)=23+2×223×2=8+86=10

So, the graphs of f and g intersect at the points (3,0), (0,0) and (2,10). These three points give two intervals of integration.

Sketch the graph of two functions as follows:

From the graph, g(x)f(x) for all x in the interval [3,0] and f(x)g(x) for all x in the interval [0,2]

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