Suppose $f'(x)=p sin(1/2x)$ where $p$ is a constant, $f(0)=1$ and $f(2pi)=0$ . Find p and hence find f(x).?

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Answer 1 · 2018-02-11T19:03:15

$p=-1/4, and, f(x)=1/2{cos(x/2)+1}$ .

$f'(x)=p*sin(x/2), (p" const.)"$

By the Definition of Integral, $f(x)=intf'(x)dx+c$ ,

$=intp*sin(x/2)dx+c$ ,

$=p*intsin(x/2)dx+c$ ,

$=p{-cos(x/2)/(1/2)}+c$ .

$rArr f(x)=-2pcos(x/2)+c.............(ast)$ .

Given that, $f(0)=1, (ast) rArr 1=-2pcos0+c$ .

$:. c-2p=1..................................................................(ast_1)$ .

Next, $f(2pi)=0, (ast) rArr 0=-2pcos(2pi/2)+c$ .

$:. c+2p=0.................................................................(ast_2)$ .

Solving $(ast_1) and (ast_2), c=1/2, p=-1/4$ .

$rArr f(x)=1/2{cos(x/2)+1}$ .

Suppose f'(x)=p sin(1/2x)f'(x)=p sin(1/2x) where pp is a constant, f(0)=1f(0)=1 and f(2pi)=0f(2pi)=0. Find p and hence find f(x).?