Eli5 how do we know that the integration of x^n is x^(n+1)/n+1
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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)
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