1. ## integration=anti-differentiation

Hi

I need some help to prove that integration is the opposite of differentiation.

thx

2. I think you want this.

3. When you say integration, I assume you mean finding the area bound by the function $\displaystyle f(x)$ and the $\displaystyle x$ axis, between $\displaystyle x = a$ and $\displaystyle x = b$.

Suppose that you made $\displaystyle n$ subdivisions $\displaystyle x_0, x_1, x_2, \dots , x_n$ where $\displaystyle a = x_0$ and $\displaystyle b = x_n$. Also mark the midpoint of each subdivision $\displaystyle m_i$ so that $\displaystyle x_0 < m_1 < x_1 < m_2 < x_2 < \dots < m_n < x_n$.

Then the area of each rectangle formed by the subdivisons is $\displaystyle (x_i - x_{i-1})f(m_i)$, and the entire area can be approximated by

$\displaystyle A \approx \sum_{i = 1}^{n}[(x_i - x_{i-1})f(m_i)]$.

The more subdivisions we make, the closer this approximation gets to the exact area. Therefore

$\displaystyle A = \lim_{n \to \infty}\sum_{i = 1}^{n}[(x_i - x_{i-1})f(m_i)]$.

Now we need to examine the Mean Value Theorem.

$\displaystyle f'(c) = \frac{f(b) - f(a)}{b - a}$ for some $\displaystyle c \in (a, b)$.

We can rewrite this equation as

$\displaystyle (b - a)f'(c) = f(b) - f(a)$.

Notice that the LHS is of the exact form of the summand for your area. We can use the Mean Value Theorem to simplify the summand, since we are making the subdivisions infinitessimally small so that the midpoint is the only point in between.

Therefore

$\displaystyle A = \lim_{n \to \infty}\sum_{i = 1}^n{[(x_i - x_{i - 1})f(m_i)]}$

$\displaystyle = \lim_{n \to \infty}\sum_{i = 1}^n[F(x_i) - F(x_{i - 1})]$ where $\displaystyle \frac{d}{dx}[F(x)] = f(x)$

$\displaystyle = \lim_{n \to \infty}\{[F(x_1) - F(x_0)] + [F(x_2) - F(x_1)] + [F(x_3) - F(x_2)] + \dots + [F(x_n) - F(x_{n - 1})]\}$

$\displaystyle = \lim_{n \to \infty}[F(x_n) - F(x_0)]$

$\displaystyle = \lim_{n \to \infty}[F(b) - F(a)]$

$\displaystyle = F(b) - F(a)$ since the function no longer depends on $\displaystyle n$.

Therefore the area bounded by a function $\displaystyle f(x)$ and the $\displaystyle x$ axis between $\displaystyle x = a$ and $\displaystyle x = b$ is $\displaystyle F(b) - F(a)$, where $\displaystyle F(x)$ is an antiderivative of $\displaystyle f(x)$.

4. My question would be "how are you defining 'antidifferentiation' and 'integration'"?

If you are using the usual definitions, that the "antiderivative" of a function f(x) is a function F(x) such that F'(x)= f(x) and "integration" is " $\int_a^b f(x)dx$= the area of the region bounded by the curve y= f(x), x= a, x= b. y= 0 (assuming f(x)> 0 for x between a and b)", then you are trying to prove the "Fundamental Theorem of Calculus".

Any text book on Calculus should include that. Wikipedia has a good discussion:
Fundamental theorem of calculus - Wikipedia, the free encyclopedia

What Prove It posted is a nice outline of that proof.

5. Originally Posted by Bouga
Hi

I need some help to prove that integration is the opposite of differentiation.

thx
How far did you get?

6. @ Ackbeet , thank you very much that was very helpful
and to prove it as well.

I had a proof that I couldn't understand, which uses rieman definition and continuity of the function

But now I get it.

7. You're welcome. Have a good one!