Definite integral property

Question

I need to prove or disprove this fact

If $f$ is integrable on $[a,b]$, then there exists a number $c$ in the open interval $(a,b)$ such that

$$\int_a^c f(x) \, dx = \int_c^b f(x) \,dx$$

I have tried to use this property

$$\int_a^b f(x) \, dx = \int_a^c f(x) \,dx + \int_c^b f(x) \,dx $$

but I am not able to prove it or find a counterexample

Thanks in advance

score 2 · Answer 1 · answered Mar 06 '15 at 12:27

2

Think about the function $F(t) = \int_a^t f(x)dx$. It is continuous. Now apply the intermediate value theorem with the constant $\frac12\int_a^b f(x)dx$.

answered Mar 06 '15 at 12:27

Gregory Grant

14,874

I don't understand your answer, can you explain a little bit more please. – Usser123456798 Mar 06 '15 at 12:47
I'm not sure how to involve $ \frac{1}{2}\int_{a}^{b} f(x) dx $ in the intermediate value theorem – Usser123456798 Mar 06 '15 at 13:09
It is the constant, which is between $F(a)$ and $F(b)$. – Gregory Grant Mar 06 '15 at 14:05

score 1 · Accepted Answer · answered Mar 06 '15 at 12:29

1

For all $c \in [a,b]$, let $$ g(c) = \int_a^c f(x)dx - \int_c^b f(x) dx. $$ The function $g$ is continuous on $[a,b]$. Moreover $g(a)$ and $g(b)$ are of opposite signs. The conclusion follows from the Intermediate value theorem.

answered Mar 06 '15 at 12:29

Siméon

10,664
1
21
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Great solution, thanks – Usser123456798 Mar 06 '15 at 12:41
@Usser123456798: You're welcome. Since you're new to this website, let me tell you it is customary to upvote and/or accept useful answers. – Siméon Mar 06 '15 at 13:09

Definite integral property

2 Answers2