#### To determine

**(a)**

**To find:**

The average value of f on the given interval.

#### Answer

1

#### 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=x-32, [2, 5]

**4) Calculation:**

Here fx=x-32, a=2 and b=5

Substituting all the above in the formula,

fave=15-2∫25x-32dx

Simplify.

fave=13∫25x-32dx

Integrating,

=13x-33325

=19x-3325

=195-33-2-33

=1923--13

=198-(-1)

=199

=1

Therefore,

fave=1

**Conclusion:**

The average value of fx=x-32 on the interval [2, 5] is 1.

#### To determine

**(b)**

**To find:**

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

#### Answer

2 or 4

#### Explanation

**1) Concept:**

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

**2) Given:**

fx=x-32

**3) Calculation:**

From part (a)

fave=1

Now, fc=c-32

From the given condition,

fave=fc

Therefore,

c-32=1

c2-6c+9=1

c2-6c+8=0

By factorization,

c2-4c-2c+8=0

cc-4-2c-4=0

c-2c-4=0

⇒c-2=0 or c-4=0

⇒c=2 or c=4

**Conclusion:**

Therefore,

The value of c=2 or 4

#### 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=x-32

**3) Calculation:**

From the part (b),

fave=fc=1 for c=2 or 4

Since,

?x=x-32 is positive function, there is a number c=2 or 4 such that the rectangle with base [2, 5] and height fc=1 has the same area as the region under the graph of f from 2 to 5.

Therefore, the graph is

**Conclusion:**

Therefore,

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