#### To determine

**(a)**

**To find:**

The average value of fx=1x on 1, 4.

#### Answer

23

#### Explanation

**1) Concept:**

Use the mean value theorem for integrals.

**2) Theorem:**

If f is continuous on [a, b], then there exists a number c in [a, b] such that

fc=fave =1b-a∫abfxdx

That is,

∫abfxdx=f(c)(b-a)

**3) Given:**

fx=1x

**4) Calculations:**

fx=1x is continuous on the given interval.

Therefore,

fave =1b-a∫abfxdx

By substituting the values,

fave=14-1∫141xdx

After integrating,

fave=132x1241

Applying the Fundamental Theorem of Calculus,

fave=132412-2112

After simplifying,

fave=1322-2=23

**Conclusion:**

The average value of fx=1x on 1, 4 is 23.

#### To determine

**(b)**

**To find:**

The value of c guaranteed by the mean value theorem for integrals such that fave=fc

#### Answer

94

#### Explanation

**1) Concept:**

Use the mean value theorem for integrals.

**2) Theorem:**

If f is continuous on [a, b], then there exists a number c in [a, b] such that

fc=fave =1b-a∫abfxdx

That is,

∫abfxdx=f(c)(b-a)

**3) Given:**

fx=1x

**4) Calculations:**

fx=1x is continuous on the given interval.

Therefore, by the mean value theorem for integrals,

f(c)=fave =1b-a∫abfxdx

From part (a),

fc=23 Since,fx=1x, fc=1c

Therefore,fc=1c=23

After taking reciprocal,

c=32

Taking square root,

c=94

**Conclusion:**

The value of c guaranteed by the mean value theorem for integrals such that fave=fc is 94.

#### To determine

**(c)**

**To sketch:**

The graph of f on 1, 4 and a rectangle whosearea is the same as the area under the graph of f.

#### Answer

#### Explanation

The graph of f on 1, 4 and a rectangle whosearea is the same as the area under the graph of f.