1
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.