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?


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?

  • $\begingroup$ thank you! I did it using Rolle theorem $\endgroup$ – Gaboru Jan 11 at 13:42
  • $\begingroup$ That is exactly what I expected! Remember, identifying tricks is key to solving questions. Keep this one in mind. $\endgroup$ – астон вілла олоф мэллбэрг Jan 11 at 14:17

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