Answer

Solution

1. 1

   Define

2. 2
   $$F(x) \;=\; \int_0^x f(t)\,dt, \qquad G(x) \;=\; \sin(2x), \qquad x \in [0,\pi/4].$$
3. 3

   Both are continuous on $[0,\pi/4]$ and differentiable on $(0,\pi/4)$ with

4. 4

   $F'(x) = f(x)$ and $G'(x) = 2\cos(2x)$. Moreover $G'(x) = 2\cos(2x) \neq 0$ on

5. 5

   $(0,\pi/4)$, since $0 < 2x < \pi/2$ there.

6. 6

   By Cauchy's mean value theorem, there exists $c \in (0,\pi/4)$ such that

7. 7
   $$\frac{F(\pi/4) - F(0)}{G(\pi/4) - G(0)} \;=\; \frac{F'(c)}{G'(c)},$$
8. 8

   i.e.

9. 9
   $$\frac{\int_0^{\pi/4} f(t)\,dt}{\sin(\pi/2) - 0} \;=\; \frac{f(c)}{2\cos(2c)},$$
10. 10

    which simplifies to

11. 11
    $$f(c) \;=\; 2\cos(2c)\,\int_0^{\pi/4} f(t)\,dt.$$

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