BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 5.2, Problem 50E
To determine

To find: The interval for the integral function 02f(x)dx lies between absolute minimum (m) and maximum value (M).

To identify: The property of the integral used to obtain the solution.

Expert Solution

Answer to Problem 50E

The value of the integral function lies between 2m and 2M.

The “comparison property 8 of integral” is used to calculate the value of integral.

Explanation of Solution

Given information:

The integral function is 02f(x)dx.

Show comparison property 8 of integrals:

If mf(x)M for axb, then m(ba)abf(x)dxM(ba) (1)

Calculate:

Consider the function f(x).

The function f(x) has lower limit a=0 and upper limit b=2.

The integral function has absolute maximum value as M and absolute minimum value as m.

Modify the function f(x) as shown below:

mf(x)Mfor0x2 (2)

Compare Equation (2) with Equation (1).

The value of the function lies between its absolute maximum and absolute minimum value within limits [0,2].

Thus, the property of integral “comparison property 8” is satisfied.

Therefore, evaluate the integral 02f(x)dx using “comparison Property 8 of integral”.

The expression to find the value of the integral by using Property 8 of integral as shown below:

Substitute 0 for a and 2 for b in Equation (1).

m(20)02f(x)dxM(20)2m02f(x)dx2M

Conclusion:

Compare Equation (2) with Equation (1).

The value of the function lies between its absolute maximum and absolute minimum value within limits [0,2].

Thus, the condition to apply comparison property 8 of integral is satisfied.

Therefore, the “comparison Property 8 of integral” is used to evaluate function 02f(x)dx.

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