Finding the value of a function at zero satisfying a certain differential equation.

Question

$f'(x)=\sin(x)$, $f(\pi/2)= 3/2$, $f(0)=?$

If I integrate the function with $\sin x$ I got a result of $\cos x$. What should I do next?

Answer 1

$$f'(x) = \sin(x) \longrightarrow f(x) = \int \sin(x)\ \text{d}x = -\cos(x) + C$$

Now $$f\left(\frac{\pi}{2}\right) = -\cos\left(\frac{\pi}{2}\right) + C = \frac{3}{2}$$

Since $\cos(\pi/2) = 0$ you get $C = 3/2$.

This will help you with $f(0)$:

$$f(0) = -\cos(0) + C = -1 + \frac{3}{2} = \frac{1}{2}$$

Answer 2

Note that if $f'(x)=\sin(x)$, then $f(x)=-\cos(x)+C$ for some $C\in\mathbb{R}$. To determine $C$, we plug $\frac{\pi}{2}$. So we have:$$\frac{3}{2}=f(\frac{\pi}{2})=\cos(\frac{\pi}{2})+C=C$$ meaning $f(x)=-\cos(x)+\frac{3}{2}$. Plugging $0$, we see that $f(0)=-\cos(0)+\frac{3}{2}=-1+\frac{3}{2}=\frac{1}{2}$.