Chapter 4.P, Problem 1P

If $x\sin x\pi x={\displaystyle {\int }_0^{x^2}f\left(t\right)dt\text{,}}$ where f is a continuous function, find f(4).

To determine

To find:

$f\left(4\right)$

Answer to Problem 1P

Solution:

$\frac{\pi }{2}$

Explanation of Solution

1) Concept:

The Fundamental Theorem of Calculus

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

2) Given:

$xsin\left(\pi x\right)={\int }_0^{{\mathrm{x}}^2}f\left(t\right)\mathrm{d}\mathrm{t}$

3) Calculation:

Here,

$xsin\left(\pi x\right)={\int }_0^{{\mathrm{x}}^2}f\left(t\right)\mathrm{d}\mathrm{t}$

Substitute $x^2=u$, $(x=\sqrt{u})$

$\sqrt{u} sin\left(\pi \sqrt{u}\right)={\int }_0^{\mathrm{u}}f\left(t\right)\mathrm{d}\mathrm{t}$

Differentiate with respect to $u$:

$\frac{1}{2\sqrt{u}}sin\left(\pi \sqrt{u}\right)+\sqrt{u} cos\left(\pi \sqrt{u}\right)·\frac{\pi }{2\sqrt{u}}=\frac{d}{du}{\int }_0^{\mathrm{u}}f\left(t\right)\mathrm{d}\mathrm{t}$

By using Fundamental Theorem of Calculus,

$\frac{1}{2\sqrt{u}}sin\left(\pi \sqrt{u}\right)+\sqrt{u} cos\left(\pi \sqrt{u}\right)·\frac{\pi }{2\sqrt{u}}=f\left(u\right)$

Tofind $f\left(4\right)$, substitute $u=4$

$f\left(4\right)=\frac{1}{2\sqrt{4}}sin\left(\pi \sqrt{4}\right)+\sqrt{4} cos\left(\pi \sqrt{4}\right)·\frac{\pi }{2\sqrt{4}}$

$=\frac{1}{4}sin\left(2\pi \right)+2 cos\left(2\pi \right)·\frac{\pi }{4}$

Substitute $sin\left(2\pi \right)=0 \& cos\left(2\pi \right)=1$

$=\frac{1}{4}\left(0\right)+2 \left(1\right)·\frac{\pi }{4}$

$=\frac{\pi }{2}$

Conclusion:

$f\left(4\right)=\frac{\pi }{2}$