# Thread: integration proof using MVT

1. ## integration proof using MVT

I have been asked to prove the following:

Suppose that f is continuous on [a,b]. Prove that there exists a point c such that the integral from a to be is f(c)(b-a).

Proof:
Let F(x)= integral from a to x of f(t)dt, then by the Mean Value Theorem there exists a value f(b) such that a$\displaystyle \leq$b$\displaystyle \leq$x. Furthermore, the integral from a to b of f(t)dt can be calculated to be f(c)(b-a).

This is all I have so far, I realize there are holes. Can you please help me with this?

2. ## Re: integration proof using MVT

Originally Posted by CountingPenguins
Suppose that f is continuous on [a,b]. Prove that there exists a point c such that the integral from a to be is f(c)(b-a).
Define $\displaystyle F(x) = \int_a^x {f(t)dt}$, then the function $\displaystyle F$ is continuous on $\displaystyle [a,b]$.

Moreover $\displaystyle F'(x) = f(x)$. So apply MVT to $\displaystyle F(x)$ on $\displaystyle [a,b].$

3. ## Re: integration proof using MVT

I would prove it this way, by distinguishing tree cases:
$\displaystyle a<b$
$\displaystyle a=b$
$\displaystyle a>b$
If you find $\displaystyle a<b$ then you'll find the other two, you may consider the proof for $\displaystyle a=b$ as trivial.

4. ## Re: integration proof using MVT

Originally Posted by CountingPenguins
I have been asked to prove the following:

Suppose that f is continuous on [a,b]. Prove that there exists a point c such that the integral from a to be is f(c)(b-a).

Proof:
Let F(x)= integral from a to x of f(t)dt, then by the Mean Value Theorem there exists a value f(b) such that a$\displaystyle \leq$b$\displaystyle \leq$x. Furthermore, the integral from a to b of f(t)dt can be calculated to be f(c)(b-a).

This is all I have so far, I realize there are holes. Can you please help me with this?

A different approach...

Let $\displaystyle f(x)$ be integrable function on $\displaystyle [a,b]$, then we can easily prove the following:

i)If $\displaystyle f(x)\geq m$ for all $\displaystyle x\in[a,b]$ then $\displaystyle \int_{a}^{b}f(x) \;dx \geq m(b-a)$

Proof:

For every partition $\displaystyle P$ of $\displaystyle [a,b]$ and for chosen points, $\displaystyle t_i\in\Delta x_i$ we eill have:

$\displaystyle \sigma _P(t_1,...,t_n)=\sum_{i=1}^{n}f(t_i)\Delta x_i\geq \sum_{i=1}^{n}m\Delta x_i=m(b-a)$.

$\displaystyle \int_{a}^{b}f(x) \;dx =\lim_{\Delta(P)\to 0}\sigma _P(t_1,...,t_n)\geq m(b-a)$.

ii) If $\displaystyle f(x)\leq M$ for all $\displaystyle x\in[a,b]$ then $\displaystyle \int_{a}^{b}f(x) \;dx \leq M(b-a)$

The proof is identical to proof in i).

Now, by combining i) and ii) we will get the next result:

If $\displaystyle f(x)$ is integrable on $\displaystyle [a,b]$ , and let $\displaystyle M$ be the supremum of $\displaystyle f(x)$ on $\displaystyle [a,b]$ and $\displaystyle m$ is the infinum of $\displaystyle f(x)$ on $\displaystyle [a,b]$. Then:

$\displaystyle m(b-a)\leq \int_{a}^{b}f(x) \;dx\leq M(b-a)$ (1)

By dividing (1) by $\displaystyle (b-a)$, we will get:
$\displaystyle m\leq \frac{1}{b-a}\int_{a}^{b}f(x) \;dx\leq M$
Now, using MVT we conclude the existence of $\displaystyle c\in[a,b]$ such that:
$\displaystyle f(c)=\frac{1}{b-a}\int_{a}^{b}f(x)$