# How can I prove the following integral equality?

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 $$a and $$F(c)=\int_a^btf(t)dt.$$