Eli5 how do we know that the integration of x^n is x^(n+1)/n+1
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The *antiderivative* of x^n is any function x^(n+1)/(n+1)+c, because the derivative of x^(n+1)/(n+1)+c is x^n.
The reason the derivative of x^(n+1)/(n+1)+c is x^n is [binomial theorem](https://en.wikipedia.org/wiki/Power_rule#Proof_by_binomial_theorem_(natural_numbers)).
A simple way it goes the other way
Let m=n+1, it makes the text simple to read.
(x^m)´ = m x^m .
It is well known and you can find the proof at [https://en.wikipedia.org/wiki/Power_rule](https://en.wikipedia.org/wiki/Power_rule)
We like to remove the initial m in m x^m Multiplying with a constant has not effect (mx)´ =m (x´) so 1/m * (x^m)´ = 1/m * m x^m = x^m
We can now replace m with n+1 and we have (((x^(n+1))/(n+1)) ´ = x^(n+1)
An integral is the reverse of derivation so we get that the integral of x^(n+1) = ((x^(n+1))/(n+1) + C
C it the constant that is often forgotten we you integrate. The derivate of a constant is zero C`= 0 so w could have added that it the derivation calculation too
Integration is the area under a curve. So, one method of finding the integral of an equation is to break it up into smaller pieces to determine the area of each piece and add them together. This is called the “Rieman Sum.” And if you have any curve other than a straight line (y=mx+b), it will not be completely accurate for finitely wide areas. So we take the limit as the number of areas increases to infinitity, or the width of the areas approaches 0 (same thing in this case). However, there is an easier way.
The fundamental theorum of calculus.
Since an integral is the area under a curve, and the derivative is the rate at which a curve changes, then if we plot the rate at which a curve changes, the area under the plot of the rate is equal to the original curve.
In other words: if y=f(x) and dy=f'(x)dx, then the integral of f'(x)dx is equal to f(x) and therefore y.
So let’s prove the derivative of x^n instead.
The derivative is the limit of (f(x+h)-f(x))/h as h approaches 0, so if we plug f(x)=x^n to that equation, we get the equation f'(x)=((x+h)^n -x^n )/h and whenever we have a value of (x+h)^n it always comes out to a variation of x^n +nx^(n-1) h+(…)+nxh^(n-1) +h^n . The number of terms depends on the value of h, but it always has n+1 terms, and the value right after the first is always multiplied by n.
So now our equation is x^n +nx^(n-1) h+(…)-x^n all over h. The two x^n values will be canceled, and the nx^(n-1) h over h will reduce to nx^(n-1), and when h is equal to 0, the rest of the terms will also be zero.
So now we have proven that if f(x)=x^n then f'(x)=nx^(n-1)
So, how does that help us with your initial question?
Well, let’s sub in the variable m=(n-1). Our new equation is f(x)=x^(m+1) and f'(x)=(m+1)x^m
So we can rewrite this using the funadmental theorum of calculus to say that x^(m+1) =INTEGRAL((m+1)x^m dx)
Since (m+1) is a constant, we can take it out of the integral x^(m+1) =(m+1)INTEGRAL(x^m dx)
Divide both sides by (m+1), and we have your proof x^(m+1) /(m+1)=INTEGRAL(x^m dx)
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!
A small note, this pattern was discovered before calculus was invented. Several people independently discovered it although they couldn’t make a rigorous proof they started using it for positive whole number powers.
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Because that’s the definition.
X^1 = X
X^2 = X^(2-1) •X
-> divide both sides with X and you get what you asked for. You can prove it through induction if you want, but that wouldn’t be ELI5
You could also ask why nX-X=(n-1)X
The integration of x^n involves finding the antiderivative, or the function whose derivative is x^n.
To derive the formula for the antiderivative of x^n, we use the power rule of differentiation. This rule states that the derivative of x^n is n*x^(n-1).
Using this rule, we can work backwards and say that the antiderivative of x^n must be x^(n+1)/(n+1), since the derivative of this function is (n+1)*x^(n)/(n+1) = x^n.
Therefore, we know that the integration of x^n is x^(n+1)/(n+1).
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