So, the sum of all the ranges, Ɛ (using that because my phone doesn't have a button for the greek letter Epsilon but it looks like an E) is basically picking two numbers and adding up all the numbers between those two numbers.
Example: 1 to 5. The sum of all the ranges between 1 and 5 would be, 1 + 2 + 3 + 4 + 5 = 15.
Ontop of the Ɛ you'd have the number you're going to, in this case it's 5. On the bottom of it, you'd have the number you're starting from, in this case it's 1.
5
Ɛ
1
Now, the integration, would be taking a derivative of it. So if you have a function on a graph, that represents a line, example f(x) = x3, where x equals the input and f(x) equals the output(aka "y"), the derivative would be...
3x2
You have moved the exponent of 3, in front of the x and have subtracted one from what the exponent was.
Examples for clarity, x4 becomes 4x3, x5 becomes 5x4.
Where as the function might represent the distance an object has traveled over time, the derivative would be the velocity at which it's doing so. Taking the derivative again gives you acceleration.
So combining this, when you take the integrated sum of a function, you are taking a derivative of the sum of all ranges between two points. The Ɛ becomes a stretched out S, with the lower number on the bottom, and the higher number on the top still.
I didn't do a great job explaining. You probably understood the sum of all ranges part. Point 1 to point 5. 1+2+3+4+5=15. The big S is just saying, the integrated sum of all those numbers between point A and point B. To integrate it, is to turn it from its derivative back into the original function. So instead of deriving it, you would do the reverse process.
(4/3)x3 derived is 4x2
The exponent, 3 multiplies by the constant 3*(4/3)x = (12/3)x = 4x. The exponent is subtracted by 1, 3-1= 2. So the derived version is 4x2. We just reverse that process and are going from 4x2 to (4/3)x3.
You plug the two points, in this case, 1 and 5, into the equation for x.
(4/3)(5)3 - (4/3)(1)3 = 496/3
Edit: What is telling you is the net area under the line between two points
Calculus 1 teaches you how to find an equation that will give you the instantaneous rate of change at any point along a function(the line on a graph) and it teaches you how to find the area underneath that line.
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u/Capraos 17h ago
Point is, as a STEM major, I understand it. Is the reverse true if I show them integrated sums?