4

If $f:[a,b]\longrightarrow \mathbf{R}$ is integrable prove that there is $c\in[a,b]$ such that $\int_a^c f(t)dt = \int_c^b f(t)dt$. I set $g(x)=\int_a^x f(t)dt$ but I don't know how I must continue.

Martin Sleziak
  • 53,687
user100002
  • 61
  • 4
    What the value of $g(x)$ at $x = a$? What's the value of $g(x)$ at $x = b$? What's the value you want to have for $g(c)$? You might do better to let $h(x) = \int_a^x f(t) ~dt - \int_x^b f(t) ~dt$ and answer the same questions for $h$, but that's a matter of taste. – John Hughes Dec 10 '13 at 15:36
  • This is question 600075 ... – DonAntonio Dec 10 '13 at 16:10
  • 1
    Please, try to make the title of your question more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. – Martin Sleziak Dec 11 '13 at 17:20

1 Answers1

1

If $f(x)$ is a continuous function in $[a,b]$, we may define a function $F(x)$ such that

$F(x) = \int_a^x f(y)\;dy$

$F(x)$ will be continuous in $[a,b]$ and differenciable in $(a,b)$:

$F'(x) = f(x)$

$F(a) = 0 \;\; ; \;\; F(b) = \alpha$

(Fundamental Theorem of Calculus)

By splitting the integrals, the problem becomes: show that $\,\exists \, c\in(a,b)$ such that

$F(c) = F(b) - F(c) \Leftrightarrow F(c) = \frac{1}{2}F(b)$

Since $F(c) \in[0,\alpha]$, the existence of $c$ can be directly proved by the Intermediate Value Theorem

Daniel Cunha
  • 1,735