1. Dec 21, 2013

### Jhenrique

Hello!

I was trying to understand what means:
$$\sum_{x_0}^{x_1}f(x)\Delta x$$
(when Î”x = 1 and x âˆˆ Z, ie, a "discrete integration", topic very comun in discrete calculus).

I computed the result so:
$$\sum_{1}^{4}x^2\Delta x=\left [\frac{1}{3}x^3-\frac{1}{2}x^2+\frac{1}{6}x \right ]_{1}^{4}=F(4)-F(1)=14$$
and I sketched the graphic:

However, the Maple computes the result as:
$$\sum_{1}^{4}x^2\Delta x=>\sum_{1}^{4}x^2\cdot 1=>\sum_{1}^{4}x^2=30$$
Given a different result of calculated for me. Why?

2. Dec 21, 2013

### AlephZero

Maple is calculating the sum following the conventional rules of notation for $\sum_1^4$, i.e. the sum is 1 + 4 + 9 + 16 = 30.

In discrete calculus you are only summing 3 items, not 4. That is why you got 1 + 4 + 9 = 14.

Maybe there is an option in Maple to use discrete calculus notation, but I don't know about that.

3. Dec 24, 2013

### HallsofIvy

Staff Emeritus
This is not a very good notation! It says you are summing from $x_0$ to $x_1$ but does NOT say what step you are using- what $\Delt x[/itex is. You appear to be assuming that [itex]x_0= 1$ and $x_1= 4[/itex[ are the only values used- that is, that [itex]\Delta x= 4- 1= 3. But this is a sum, not a difference. It is [itex](1)^2(3)+ (2)^2+ 3= 3+ 12= 15$.

That graphic shows you using $\Delta x= 1$, so that x takes on values of 1, 2, and 3:
$1^2(1)+ 2^2(1)+ 3^2(1)= 1+ 4+ 9= 14$

However, the Maple computes the result as:
$$\sum_{1}^{4}x^2\Delta x=>\sum_{1}^{4}x^2\cdot 1=>\sum_{1}^{4}x^2=30$$
Given a different result of calculated for me. Why?[/QUOTE]
That sum would NOT approximate the integral from 1 to 4 because it includes a "rectangle" between x= 4 and x= 5: 1+ 4+ 9+ 16= 30.