Chapter 6.4, Problem 62E

To determine

Evaluate the function at $x=4$ .

Answer to Problem 62E

Value of function:

  $f\left(4\right)=1$

Explanation of Solution

Given information:

  ${\displaystyle {\int }_0^{x}f\left(t\right)dt}=x\cos \pi x$

Calculation:

According to FTC (Fundamental Theorem of Calculus):

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

Substitute in

  ${\displaystyle {\int }_0^{x}f\left(t\right)dt}=x\cos \pi x$

That gives

  $f\left(x\right)=\frac{d}{dx}\left(x\cos \pi x\right)$

Use product rule to differentiate:

  $f\left(x\right)=1\cdot \cos \pi x+x\cdot \left(-\sin \pi x\right)\cdot \pi =\cos \pi x-\pi x\sin \pi x$

Then

Evaluate the function at $x=4$ :

  $f\left(4\right)=1\cdot \cos \left(\pi \cdot 4\right)-4\cdot \left(\sin \left(\pi \cdot 4\right)\right)\cdot \pi =\cos 4\pi -\pi \left(4\right)\sin 4\pi =1$

Therefore,

  $f\left(4\right)=1$