#### To determine

**(a)**

**To find:**

The average value of f on the given interval.

#### Answer

43

#### Explanation

**1) Concept:**

Use the formula for the average value of f on the interval [a, b].

**2) Formula:**

fave=1b-a∫abf(x)dx

**3) Given:**

fx=x4-x2, [0, 2]

**4) Calculation:**

Here, fx=x4-x2, a=0 and b=2

Substituting all the above in the formula,

fave=12-0∫02x4-x2dx

Simplify.

fave=12∫02x4-x2dx

Here, we need to use the substitution method.

Let u=4-x2 so du=-2xdx that is xdx=-du2

Since this is a definite integral, the limits also change with respect to the substitution.

When x=0, u=4 and when x=2, u=0

Substituting all in the above integration,

fave=12∫02x4-x2dx

=12∫40u-du2

=12·12∫40- udu

Using the property of the definite integral for reversing the limit,

=14∫04u du

Simplify.

=14∫04u12 du

Integrating,

=14u323204

=212u3204

=16432-032

=168-0

=43

Therefore,

fave=43

**Conclusion:**

The average value of fx=x4-x2 on the interval [0, 2] is 43.

#### To determine

**(b)**

**To find:**

The value of c such that fave=f(c)

#### Answer

0.7135 or 1.8683

#### Explanation

**1) Concept:**

Using calculations in part (a) find the value of c

**2) Given:**

fx=x4-x2

**3) Calculation:**

From part (a)

fave=1

Now, fc=c4-c2

From the given condition,

fave=fc

Therefore,

c4-c2=43

Squaring,

c24-c2=169

4c2-c4-169=0

c4-4c2+169=0

This is quadratic in form of c2.

Using quadratic formula,

c2=-(-4)±-42-411692(1)

=4±16-6492

=4±16-6492

=4±8092

=4±23202

=2±1320

Therefore,

c2=2+1320 or 2-1320

c2=3.4894 or 0.5097

Taking square root,

c=0.714 or 1.868

**Conclusion:**

Therefore,

The value of c=0.714 or 1.868

#### To determine

**(c)**

**To sketch:**

The graph of f and rectangle whose area is same the area under the graph of f

#### Answer

#### Explanation

**1) Concept:**

The geometric interpretation of the Mean Value Theorem for Integrals is that, for positive functions f, there is a number c such that the rectangle with base [a, b] and height f(c) has the same area as the region under the graph of f from a to b

**2) Given:**

fx=x4-x2

**3) Calculation:**

From the part (b),

fave=fc=43

And c=0.714 or 1.868

Since,

?x=x4-x2 is positive function, there is a number c=0.714 or 1.868 such that the rectangle with base [0, 2] and height

fc=43

has the same area as the region under the graph of f from 0 to 2.

Therefore, the graph is

**Conclusion:**

Therefore,

The graph of f and rectangle whose area is same the area under the graph of f is