Chapter 5.P, Problem 1P

(a) Find a positive continuous function $f$ such that the area under the graph of $f$ from $0$ to $t$ is $A\left(t\right)=t^3$ for all $t>0$.

(b) A solid is generated by rotating about the $x$-axis the region under the curve $y=f\left(x\right)$, where $f$ is a positive function and $x\ge 0$. The volume generated by the part of the curve from $x=0$ to $x=b$ is $b^2$ for all $b>0$. Find the function $f$.

To determine

(a)

To find:

A positive continuous function $f$ such that the area under the graph of $f$ from $0 to t$ is $A\left(t\right)=t^3$ for all $t>0.$

Answer to Problem 1P

Solution:

$f\left(t\right)=3t^2 for all t>0$

Explanation of Solution

The area under the graph of $f$ from $0 to t$ is the same as ${\int }_0^{t}f\left(x\right)dx.$

This area is given as $t^3$ for all $t>0$

${\int }_0^{t}f\left(x\right)dx=t^3$

It is given that $f\left(x\right)$ is continuous.

Differentiating the above equation with respect to $t,$

$\frac{d}{dt}{\int }_0^{t}f\left(x\right)dx=\frac{d}{dt}\left(t^3\right)$

Using the Fundamental Theorem of Calculus (1),$\frac{d}{dt}{\int }_0^{t}f\left(x\right)dx=f\left(t\right)$

Thus it follows that

$f\left(t\right)=\frac{d}{dt}\left(t^3\right)=3t^2$

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

Conclusion:

$f\left(t\right)=3t^2 for all t>0$

To determine

(b)

To find:

The function $f.$

Answer to Problem 1P

Solution:

$f\left(x\right)=\sqrt{\frac{2x}{\pi }} , x\ge 0$

Explanation of Solution

The area of cross section at $x$ is

$A\left(x\right)=\pi y^2= \pi {\left(f\left(x\right)\right)}^2$

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

$V={\int }_0^{b}A\left(x\right)dx$

$V={\int }_0^b\pi {\left(f\left(x\right)\right)}^{2}dx$

Therefore,

${\int }_0^b\pi {\left(f\left(x\right)\right)}^{2}dx=b^2 , \forall b>0$

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

$\pi {\left[f\left(b\right)\right]}^2=2b$

Solving for $f\mathrm{(}b\mathrm{)},$

$f\left(b\right)=\sqrt{\frac{2b}{\pi }}$

Since $f$ is positive, the required function is

$f\left(x\right)=\sqrt{\frac{2x}{\pi }} , x\ge 0$

Conclusion:

$f\left(x\right)=\sqrt{\frac{2x}{\pi }} , x\ge 0$