#### To determine

**(a)**

**To find:**

The average value of f on the given interval.

#### Answer

32

#### Explanation

**1) Concept:**

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

**2) Formula:**

fave=1b-a∫abf(x)dx

**3) Given:**

fx=x3, [0, 8]

**4) Calculation:**

Here,fx=x3, a=0 and b=8

Substituting all the above in the formula,

fave=18-0∫08x3dx

Simplify.

fave=18∫08x3dx

=18∫08x13dx

Integrating,

=18x4/34308

=332(8)4/3-(0)4/3

=33216-0

=33216

=32

Therefore,

fave=32

**Conclusion:**

The average value of fx=x3 on the interval [0, 8] is 32.

#### To determine

**(b)**

**To find:**

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

#### Answer

278

#### Explanation

**1) Concept:**

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

**2) Given:**

fx=x3

**3) Calculation:**

From part (a)

fave=32

Now, fc=c3

From the given condition,

fave=fc

Therefore,

c3=32

Taking cube,

c=323

Therefore,

c=278

**Conclusion:**

Therefore,

The value of

c=278

#### 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=x3

**3) Calculation:**

From the part (b),

fave=fc=32 and c=278

Since,

fx=x3 is positive function, there is a number

c=278

such that the rectangle with base [0, 8] and height

fc=32

has the same area as the region under the graph of f from 0 to 8.

Therefore, the graph is

**Conclusion:**

Therefore,

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