   Chapter 6.2, Problem 45E ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

#### Solutions

Chapter
Section ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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# Area of a Region In Exercises 45-50, use the integration table in Appendix C to find the exact area of the region bounded by the graphs of the equations. Use a graphing utility to verify your results. y = 1 ( 16 − x 2 ) 3 / 2 ' y = 0 , x = − 2 , x = 2

To determine

To calculate: The exact area of region bounded by the graphs of equations y=1(16x2)3/2, y=0, x=2 and x=2.

Explanation

Given Information:

The provided equations are y=1(16x2)3/2, y=0, x=2 and x=2.

Formula used:

The formula for the integral 1(a2u2)3/2du is;

1(a2u2)3/2du=ua2a2u2+C

General power differentiation Rule:

ddx[un]=nun1dudx

Calculation:

Consider u=x.

Differentiate the considered function with respect to x using power rule of differentiation.

du=dx

Consider the provided expression:

y=1(16x2)3/2

To find the exact area of region bounded by the graphs of equations y=1(16x2)3/2, y=0, x=2 and x=2, then integrate equation y=1(16x2)3/2 with lower limit x=2 and higher limit x=2, which can be mathematically written as:

22ydx=221(16x2)3/2dx

Rewrite.

22ydx=221(42x2)3/2dx

Here, a=4, u=x and du=dx.

Substitute u for x, a for 4, and du for dx.

22ydx=221(a2u2)3/2du

Use the above formula of integration and solve the above integral as;

22ydx=[ua2a2u2]22

Substitute x for u, and 4 for a.

22ydx=[x4242x2]22=116[x16x2]22=116[(2)16(2)2(2)16(2)2]=116[2164+2164]

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