# [Math] How to work out an integral with algebra.

calculusderivativesintegration

I'm a bit of a noob and I am only just starting to learn calculus. I know how to work out a derivative with algebra (instead of using those rules and shortcuts) but I don't know how or if you can do the same with integration.

Here is the method I used to work out a derivative with algebra (and limits):

$f(x) = x^2,$

$f'(x) = \lim_{h\to 0} \frac{f(x + h) – f(x)}{h}$

$= \lim_{h\to 0} \frac{(x + h)^2 – x^2}{h}$

$= \lim_{h\to 0} \frac{x^2 + 2xh + h^2 – x^2}{h}$

$= \lim_{h\to 0} \frac{2xh + h^2}{h}$

$= \lim_{h\to 0} 2x + h$

$= 2x$

Otherwise I could just use the power rule for derivatives, but I want to know how and if I can work out the antiderivative (integral) using an algebraic method like the one above rather than using rules etc. e.g. how would I work out that x^2 is the antiderivative of 2x. I know there is methods where you just do long summations and get approximate values of the 'area under the curve' but I want to find out how and if I can get the actual antiderivative expression e.g. x^2 (in the case of 2x). I hope I am not out of my depth 🙂

Path 1: Riemann Sums

This is fairly advanced material, and the general form is usually not taught until a first year university course in calculus. Its primary importance is rigourously defining (definite) integrals - it's not considered a practical method.

But I will apply it to your function $2x$ with the bounds $0$ to $u$. First, we split up the interval $[0, u]$ into $n$ rectangles. These each have width $u\over{n}$. The goal is for the sum of the areas of these rectangles to approximate the area under the curve while there's a finite amount of them, then take a limit as the quantity $n$ goes to infinity, giving us an exact result. We take the height of each rectangle to be that of the function on, say, the left side of the rectangle.

We have one rectangle at $x = 0$, one at $x = {u\over{n}}$, one at $2u\over{n}$, one at $3u\over{n}$ and so on. Because the function is $2x$, the height of each of these rectangles is just twice their placement on the x axis: $0$, $2u\over{n}$, $4u\over{n}$, etc. Multiplying in the width of each rectangle, we get that the rectangles have areas: $0$, $2u^2\over{n^2}$, $4u^2\over{n^2}$, etc. Better stated, the total area of the rectangles is: $$\sum\limits_{k=0}^{n-1}{2k \ u^2\over{n^2}} = {2 \ u^2\over{n^2}}\sum\limits_{k=0}^{n-1}k$$

There is a nice formula for the sum of the integers up to a certain value (you can easily find analogous ones for squares, cubes and so on) which allows us to re-express that as:

$${2 \ u^2\over{n^2}} \cdot {n(n-1)\over{2}} = u^2 {n-1 \over{n}} = u^2 -{u^2\over{n}}$$

As $n$ goes to infinity that goes to $u^2$. Congratulations, you have your integral.

Path 2: Power Series

If we assume you're dealing with fairly well behaved functions - those with power series representations - then there is an approach (of sorts) to this.

For reference, a power series is basically an infinite degree polynomial. E.g. $$\sin(\theta) = \theta-\frac{\theta^3}{6}+\frac{\theta^5}{120}-\frac{\theta^7}{5040} + \cdot \cdot \cdot = \sum\limits_{k = 0}^{\infty} \frac{(-1)^k \theta^{2k+1}}{(2k+1)!}$$

Now, with the help of the binomial theorem, you can prove from the definition you used above that the derivative of $x^n$ (wrt x) is $n \ x^{n-1}$. But indefinite integration and differentiation are ~inverses, so this also proves that the integral of $x^n$ is $x^{n+1} \over{n+1}$. Armed with that knowledge and a power series representation for the function you have in mind, you can integrate anything.