# Proving a statement with the mean value theorem

by Gaboru   Last Updated January 11, 2019 12:20 PM - source

If $$f:[-2,2]\rightarrow R$$ is a continuous function, there exists $$c$$ in $$(-2,2)$$ such that $$\int_{-2}^cf(x)dx=\left( \frac{4}{c^3}-\frac{c}{4}\right) f(c)$$.I tried to prove this using the mean value theorem.

So there exists $$c$$ in $$(-2,2)$$ such that $$\int_{-2}^2f(x)dx=4f(c)$$. Can somebody tell me what should I do next, please?

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First, we define the function $$g : [-2,2] \to \mathbb R$$ by $$g(x) = \int_{-2}^x f(t)\mbox{dt}$$. By the fundamental theorem of calculus, $$g$$ is differentiable, with $$g'(x) = f(x)$$.
Consider the function $$h(x) = 16 - x^4$$ on $$[-2,2]$$. Then, note that: $$\frac{4}{c^3}- \frac c4 = \frac{16 - c^4}{4c^3} = -\frac{h(c)}{h'(c)}$$
We are essentially asked to show that there is a $$c$$ such that $$g(c) = \frac{-f(c)h(c)}{h'(c)}$$, or $$h'(c)g(c) = -g'(c)h(c)$$, or $$(gh)'(c) = 0$$ by the product rule.
Note that $$h(-2) = h(2) = 0$$. Conclude that $$gh(-2) = gh(2) = 0$$. Can you use some theorem now?