Eli5 how do we know that the integration of x^n is x^(n+1)/n+1
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We know that because integration is just the opposite of differentiation. So the question becomes why is the differentiation of x^(n+1)/(n+1) equal to x^n?
Differentiation is finding the steepness of a curve at a particular point, which is (x,y).
To do this, let’s first look at the steepness between any two points. For two points, (x1,y1) and (x2,y2), the steepness of a straight line between them is the change in height over the change in length i.e. (y2-y1)/(x2-x1)
Now let’s consider two points on the curve y=x^(n+1)/(n+1). Let’s consider the points (x,x^(n+1)/(n+1)) and (x+h,(x+h)^(n+1)/(n+1)). Then the slope of a straight line between them, the change in height over the change in length, is [(x+h)^(n+1)/(n+1) – x^(n+1)/(n+1)]/(x+h – x).
This simplifies down to 1/(n+1) * [(x+h)^(n+1) – x^(n+1)]/h
Now let’s multiply out the (x+h)^(n+1) term. There’s a mathematical theorem called the binomial theorem, which tells you how to multiply out brackets like this. The answer it gives is that this multiplies out to give x^(n+1) + (n+1)hx^n + [a load of stuff starting with h^2 or higher].
I’m gonna call that “load of stuff” O(h^2) – it’ll hopefully become clear later why.
This now makes our slope 1/(n+1) * [x^(n+1) + (n+1)hx^n + O(h^2) – x^(n+1)]/h
Cancelling out the x^(n+1) terms and dividing by the h gives us a slope of:
1/(n+1) * [(n+1)x^n + O(h^1)].
Now, when we differentiate, we want to take the slope at a specific point, i.e. the slope as x and x+h gets closer together i.e. the slope as h reduces to 0.
When h is 0, O(h^1) is also 0 (as all its terms involve a h term). So, as h reduces to 0, the slope reduces to:
1/(n+1) * (n+1)x^n = x^n.
Appreciate there’s lots of algebra involved here – there’s a reason why calculus is considered advanced maths!
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