# Thread: Error in Integration....

1. ## Error in Integration....

[NOT A MATH MAJOR]

Is there an error involved in Integration calculation? If area of a irregular shape is calculated by integration, would that be different than the actual value? The basics of integration is dividing the curve in infinite small parts and adding their individual values...

Theoretically I understand since there are infinite parts, error can not come.

Has someone formally studied this thing?

(I am not interested in numerical integration errors.)

2. Originally Posted by vacuumBubble
[NOT A MATH MAJOR]

Is there an error involved in Integration calculation? If area of a irregular shape is calculated by integration, would that be different that the actual value? The basics of integration is dividing the curve in infinite small parts and adding their individual values...

Theoretically I understand since there are infinite parts, error can not come.

Has someone formally studied this thing?

(I am not interested in numerical integration errors.)
No, if the definite integral exists, then there is not any error. This is what integral calculus was invented for...

3. Originally Posted by Prove It
No, if the definite integral exists, then there is not any error. This is what integral calculus was invented for...
I was hoping for some formal methods or proofs...

4. I would advise you to check Youtube, there are numerous demonstration videos proving this.

5. Originally Posted by Prove It
I would advise you to check Youtube, there are numerous demonstration videos proving this.
Thanks for the pointer.

I just could not find the right keywords for search.

I tried on google first, and then posted the que here.

Can you please tell me what to look for?

6. Try words like "Definite Integral", "Riemann Sum", "Proof"

7. Originally Posted by Prove It
Try words like "Definite Integral", "Riemann Sum", "Proof"

Huh, feeling dumb now.

Could not find a "proof". Videos explain the definite integral as the limit of sum, but no formal proof.

The concept either has to be an axiom or has to have a proof.

Is mathematics accepting something on the basis of Intuition or Heuristics for a change?

8. Originally Posted by vacuumBubble
Huh, feeling dumb now.

Could not find a "proof". Videos explain the definite integral as the limit of sum, but no formal proof.

The concept either has to be an axiom or has to have a proof.

Is mathematics accepting something on the basis of Intuition or Heuristics for a change?
You couldn't find a video that turns the Riemann Sum into a telescoping series by use of the Mean Value Theorem?

9. Since you say you are not a math major, what would you consider a "formal proof"?

A formal proof would require many concepts that you, perhaps, are not familiar with.

Here is a sketch of a proof that if f(x) is a continuous, integrable, function from a to b, and $f(x)\ge 0$ for x between a and b, then $\int_a^b f(x)dx$ is the area of the region bounded by x= a, x= b, y= 0, and y= f(x).

Divide the interval from a to b on the x-axis into n equal length sub intervals. Let $\Delta x$ be the length of each interval. Since f is continuous, there exist, in each sub interval, a point, $x^*$, such that $f(x^*)$ is maximum on that sub interval and a point, $x_*$, such that $f(x_*)$ is minimum on that sub interval.

Construct two rectangles on each interval, having $\Delta x$ as base and $f(x^*)$ as height for one and $f(x_*)$ as height for the other. The area of the larger rectangle is $f(x^*)\Delta x$ and the area of the smaller rectangle is $f(x_*)\Delta x$. Further, since those are max and min values of f, the graph of y= f(x) lies between the tops of the two rectangles and $f(x_*)\Delta x\le A_i\le f(x^*)\Delta x$.

The area under all of the larger rectangles is the sum of the areas of the individual rectangles, $\sum f(x^*)\Delta x$, and the area of all the small rectangles is $\sum f(x_*)\Delta x$. Since the graph itself always lies between the tops of the two rectangles in every sub interval, we must have
$\sum f(x_*)\Delta x\le A\le \sum f(x^*)\Delta x$
where, now, A is the area under the graph.

What happens as we take n larger and larger, that is, take more and more subintervals? By definition of "integrable", the limit of those two sums, as n, the number of subintervals, goes to infinity, must be the same- and we call that common value the definite integral- $\int_a^b f(x) dx$.

(This is the crucial point- it is the limit concept that removes the "errors" involved in each step.)

And, since A, the area of that region, is always between the two sums, no matter what n is, it must be equal to the common value: $A= \int_a^b f(x) dx$.

Of course, there are technical details. For one, you have to define area for an oddly shaped region in such a way that
1) If A is a subset of B, then Area(A) is less than or equal to Area(B).
2) If A and B have no common points except possibly on their boundary, then $Area(A)\cup Area(B)= Area(A)+ Area(B)$
and, of course,
3) Area of a rectangle with height h and width w is hw

so that all of the things we say in the proof are valid.

You may think it is cheating to require from the start that the function be "integrable" but there are functions that are not integrable- there exist sets which do NOT have "area" (not even 0 area) .

You can show that if a function is both continuous and bounded, then it is integrable- but that's a different proof.

10. An alternative proof:

First of all, you need to know of the Mean Value Theorem, which states that on a continuous function $f(x)$ between $x = a$ and $x = b$, the secant made between these points will have the same gradient as some point in between $a$ and $b$, call it $c$.

In other words:

$\frac{F(b) - F(a)}{b - a} = f(c)$

or $F(b) - F(a) = f(c)[b - a]$.

It is important to note that this is linking a function with its derivative (from the point of view of $F$), or linking a function with its antiderivative (from the point of view of $f$).

In other words, the derivative of $F$ is $f$, or the antiderivative of $f$ is $F$.

Now, suppose that we had some function that is positive and continuous between $x_0$ and $x_n$.

It can be divided into subintervals along the $x$ axis, such that

$x_0 < x_1 < x_2 < \dots < x_n$.

We can also define point $t_i$ in between each $x_i$ and $x_{i + 1}$.

In other words

$x_0 < t_0 < x_1 < t_1 < x_2 < t_2 < \dots < t_{n - 1} < x_n$.

By doing this, we have roughly split up the region into some rectangles of length $x_{i + 1} - x_i$, with height $f(t_i)$.

Therefore, the area of each rectangle is $f(t_i)[x_{i + 1} - x_i]$

And the area of the region is

$A = \sum_{i = 0}^{n - 1} f(t_i)[x_{i + 1} - x_i]$.

Now if we go back to our Mean Value Theorem:

We have $f(c)[b - a] = F(b) - F(a)$, where $a < c < b$.

Can you see that if we set $a = x_i, b = x_{i + 1}, f(c) = f(t_i)$ then we have

$f(t_i)[x_{i + 1} - x_i] = F(x_{i + 1}) - F(x_i)$.

That means our Area is

$A = \sum_{i = 0}^{n - 1} f(t_i)[x_{i + 1} - x_i]$

$= \sum_{i = 0}^{n - 1}[F(x_{i + 1}) - F(x_i)]$.

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

Can you see how the middle terms are cancelling each other out? This is called a telescopic series, and will evaluate to

$= F(x_n) - F(x_0)$.

Notice that we had a little bit of error as far as the rectangles go, since we don't know exactly how high they are. This is because we only specified $t_i$ to be somewhere in between $x_i$ and $x_{i + 1}$ and so we didn't know exactly the height $f(t_i)$. But as long as we increase the number of subintervals, in other words, make $n \to \infty$, the width of each subinterval becomes the same as the point $t_i$ and thus $f(t_i)$ CAN be evaluated.

So the definite integral IS accurate, because as we increase the number of subdivisions, the rectangles converge upon the area of the region (and the middle terms end up vanishing through the telescopic series anyway).