Let $f:[a,b] \to \Bbb R $ be a continuous function. Prove that there's a c, $c \in (a,b)$ with the following property: $$\int_a^btf(t)dt=a\int_a^cf(t)dt+b\int_c^bf(t)dt$$ Thanks in advance.


I will use a lemma that I hope you are familiar with. If not, you can try to prove it, it's pretty easy.

Lemma: if $f:[a,b]\to\mathbb{R}$ is continuous then there is a point $c\in (a,b)$ such that $\int_a^b f(x)dx=f(c)(b-a)$.

Alright, now I'll show how to prove your problem. Let $F(x)=\int_a^x f(t)dt$. Since $f$ is continuous we know by the fundamental theorem of calculus that $F'(x)=f(x)$. So now from integration by parts we get:

$\int_a^b tf(t)dt=tF(t)|_a^b-\int_a^b F(t)dt=bF(b)-aF(a)-\int_a^b F(t)dt=bF(b)-F(c)(b-a)=$

$aF(c)+b(F(b)-F(c))=a\int_a^c f(t)dt+b\int_c^b f(t)dt$

I used the lemma I stated and also the trivial facts that $F(a)=0$ and $F(b)-F(c)=\int_c^b f(t)dt$.


We know that $$a\int_a^bf(t)dt=\int_a^baf(t)dt < \int_a^btf(t)dt < \int_a^bbf(t)dt = b\int_a^bf(t)dt.$$

Let $F(x)= a\int_a^xf(t)dt+b\int_x^bf(t)dt$ for $x\in [a,b]$. Note that $F(a)=b\int_a^bf(t)dt$, $F(b)=a\int_a^bf(t)dt$, and $F(x)$ is continuous. Therefore, by IVT, there exists some $c$ such that $a<c<b$ and $F(c)=\int_a^btf(t)dt.$


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.