   Chapter 6.2, Problem 46E ### 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
1 views

# 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 ( x 2 + 0.25 ) 3 / 2 ' y = 0 , x = 0 , x = 1

To determine

To calculate: The exact area of region bounded by the graphs of equations y=1(x2+0.25)3/2, y=0, x=0 and x=1.

Explanation

Given Information:

The provided equations are y=1(x2+0.25)3/2, y=0, x=0 and x=1.

Formula used:

The formula for the integral 1(u2±a2)3/2du is;

1(u2±a2)3/2du=±ua2u2±a2+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(x2+0.25)3/2

To find the exact area of region bounded by the graphs of equations y=1(x2+0.25)3/2, y=0, x=0 and x=1, then integrate equation y=1(x2+0.25)3/2 with lower limit x=0 and higher limit x=1, which can be mathematically calculated as;

01ydx=011(x2+0.25)3/2dx

Rewrite.

01ydx=011(x2+0.52)3/2dx

Here, a=0.5, u=x and du=dx.

Substitute u for x, a for 0.5, and du for dx;

01ydx=011(u2+a2)3/2du

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

01ydx=[ua2u2+a2]01

Substitute x for u, and 0.5 for a.

01ydx=[x0

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started 