#### To determine

**(a)**

**To find:**

A positive continuous function f such that the area under the graph of f from 0 to t is At=t3 for all t>0.

#### Answer

ft=3t2 for all t>0

#### Explanation

The area under the graph of f from 0 to t is the same as ∫0tfxdx.

This area is given as t3 for all t>0

∫0tfxdx=t3

It is given that fx is continuous.

Differentiating the above equation with respect to t,

ddt∫0tfxdx=ddtt3

Using the Fundamental Theorem of Calculus (1),ddt∫0tfxdx=f(t)

Thus it follows that

ft=ddtt3=3t2

This is the required positive continuous function for all t>0

**Conclusion:**

ft=3t2 for all t>0

#### To determine

**(b)**

**To find:**

The function f.

#### Answer

fx=2xπ , x≥0

#### Explanation

The area of cross section at x is

Ax=πy2= πfx2

The volume of the solid between x=0 to x=b is

V=∫0bAxdx

V=∫0bπfx2dx

Therefore,

∫0bπfx2dx=b2 , ∀ b>0

Differentiating the above equation with respect to b using the Fundamental Theorem of Calculus,

πfb2=2b

Solving for f(b),

fb=2bπ

Since f is positive, the required function is

fx=2xπ , x≥0

**Conclusion:**

fx=2xπ , x≥0