Let's try to take the integral of y=2x.y=2x. We'll find the area under the line from 00 to some arbritary point x,x, just as we found the slope at some point xx during differentiation.

First, we visualize our function - we have a triangle, where the yy value is always twice the xx value. So, if we take the first vertical bar with a width of dx,dx, then the height of that bar would be 2dx.2dx. And, if we took the next vertical bar with width dx,dx, then our distance from the origin is now 2dx,2dx, so we can raise the bar as high as 4dx.4dx. And, if we take the third bar with width dx,dx, then we are now 3dx3dx out from the origin, so we can raise the bar as high as 6dx.6dx.

Now that we have the dimensions of the first few bars, let's see what happens when we find their area:


And if we take the next few bars, we would have


How long does this series continue for? That would depend on how many bars there are.

Let's see: if the total width of the area is x,x, and the width of each bar is dx,dx, then there should be xdx\frac{x}{dx} bars. So the complete series would look like:


How do we find the sum of this series? We can use the Gauss method:


After simplification:


Now, we get to decide how wide our bars should be - i.e., the value of dx.dx. Well, the smaller our bars are, the more accurate our calculation will be. So, we're going to go with infinitesimally small:


We are done! We can now write our answer as

2x dx = x2+C.\int2x\ dx\ =\ x^2+C.

Why, you may ask, do we have to write " +C+C "? What is CC ?

Before we talk about that, let's ask ourselves another, explorable question - why is it, that:


for some constant C,C, and

2x dx =x2+C?\int2x\ dx\ =x^2+C?

Well, let's think about it- when we take the differentiation of the integration of something, we are asking ourselves how much the integration changes when we increase the upper bound by dx.dx.

This is a question we know how to answer, because dxdx is the width of one of our rectangles! Each time we increase the upper bound, we are adding another rectangle. Every time that we add another rectangle to the end, the height of that rectangle is going to be twice our upper bound, since y=2x.y=2x. And our if our upper bound is x,x, (because we are trying to find a general formula here), then the area of that rectangle is going to be 2x dx.2x\ dx.

Wait a minute - isn't that the function that we were trying to integrate? Yes! This is no coincidence - it's the Fundamental Theorem of Calculus.

We'll look deeper at the actual theorem and the meaning of " +C+C " in the next lesson. See you there!

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