Chapter 6, Problem 1P

To determine

To find : The function $f$ .

Answer to Problem 1P

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

Explanation of Solution

Given information :

The volume is generated by the part of the curve $y=f\left(x\right)$ from $x=0$ to $x=b$ is $b^2$ for all $b>0$ .

Concept used :

The area under the graph of $f$ from $0$ to $t$ is equal to ${\displaystyle \underset{0}{\overset{t}{\int }}f\left(x\right)}dx$ .

If, ${\displaystyle \underset{0}{\overset{t}{\int }}f\left(x\right)}dx=t^3$

We differentiate with both sides with respect $t$ we get,

  $f\left(t\right)=3t^2$ [Using fundamental theorem of calculus.]

Calculation :

Now generated volume from $x=0$ to $x=b$

  $={\displaystyle \underset{0}{\overset{b}{\int }}\pi {\left[f\left(x\right)\right]}^2}dx$

By given information ${\displaystyle \underset{0}{\overset{b}{\int }}\pi {\left[f\left(x\right)\right]}^2}dx=b^2\text{ for all }b>0$

Differentiating both sides with respect to $b$ we get,

  $\begin{array}{l}\pi {\left[f\left(b\right)\right]}^2=2b\text{ [using fundamental theorem of calculus]}\\ \Rightarrow f\left(b\right)=\sqrt{\frac{2b}{\pi }}\Rightarrow f\left(x\right)=\sqrt{\frac{2x}{\pi }}\end{array}$