# Definite Integrals

In the previous article, we discussed the idea of finding the area under a curve by adding rectangles. At the end of it, I presented a formula which would represent the area under the curve as: $lim_{n \to \infty} \sum_{i=1}^{n} f(x_i) \Delta x$

As it stands, this formula may very well tell us the area under a curve, but we have no good way to solve it, aside from manually doing lots of calculation. Our next goal is to find an elegant way to solve this equation for any function we are given. A lot of textbooks immediately introduce the idea of a definite integral, but I want to take some time to show the intuition of where it comes from.

Suppose we had a function A(x) which could tell us the area under the curve for some function f(x), at any x value. Think about what it would mean to take the derivative of A(x). If we take the derivative, we are finding the rate that the area changes over time. The area of the curve is changing based on the length and height of the curve itself. In other words, the change in the area is equivalent to the change in the y position based on the change of the x position.

If this sounds familiar, it is because we just described the definition of a derivative. This tells us that A'(x) = f(x). This relationship is very significant, as it gives us a relationship between differential and integral calculus. Recall that we discussed the idea of antiderivatives, which was an operation that found a function whose derivative is the function we started with. It stands that if A'(x) = f(x), then the antiderivative of f(x) is A(x). This means that to find the area under a curve f(x), we simply need to find the antiderivative of f(x).

We will refer to this antiderivative idea as an integral. Specifically, a definite integral is an antiderivative that is evaluated at two points, to find the area under a curve between two points. It is defined as follows:

$\int_{a}^{b}f(x) dx = lim_{n \to \infty} \sum_{i=1}^{n} f(x_i) \Delta x = A(x)$

We introduce a new symbol, known as the integral symbol, which is the

Since this definition is based on limits, we can conclude a few properties for our definite integral that are useful.

**Properties of definite integrals**

- $\int_{a}^{b} c dx = c(b-a)$, where c is any constant
- $\int_{a}^{b} [f(x)+g(x)]dx = \int_{a}^{b} f(x)dx + \int_{a}^{b}g(x)dx$
- $\int_{a}^{b} cf(x)dx = c \int_{a}^{b}f(x)dx$, where c is any constant
- $\int_{a}^{b} [f(x) - g(x)] dx = \int_{a}^{b} f(x) dx - \int_{a}^{b} g(x) dx$

I'll show how to prove one of these properties, since they are all proven in a similar way.

**Proof of property 3**

Let's prove that property 3, $\int_{a}^{b} cf(x)dx = c \int_{a}^{b}f(x)dx$, is true.

To do this, we will use the summation notation definition for a definite integral.

$\int_{a}^{b} cf(x)dx = lim_{n \to \infty} \sum_{i=1}^{n}c*f(x_i) \Delta x$

$= c * lim_{n \to \infty} \sum_{i = 1}^{n} f(x_i) \Delta x$

$= c * \int_{a}^{b} f(x) dx$

We can factor out the constant due to limit laws. From our limit laws, we know that $lim_{x \to a} c*f(x) = c * lim_{x \to a} f(x)$

There is one other property of integrals we can discuss that is helpful to know.

**Property: **$\int_{a}^{c} f(x) dx + \int_{b}^{c} f(x) dx = \int_{a}^{b}f(x) dx$

This simply tells us that we can combine integrals, such that the end of one is equal to the start of the next one. With these properties, we are able to manipulate integrals into more useable forms to solve. One thing we haven't really discussed so far is how we even find antiderivatives in the first place. In the article on antiderivatives, we solved them through inspection, which works in some cases. There are actually a number of strategies used to solve integrals, that we will focus on for most of integral calculus. In general, taking an integral is much harder compared to taking a derivative, and we will see cases where the best we can do is estimate the value. These situations are not ideal, however we can most of the time get a pretty good estimate using techniques similar to the rectangle summations we discussed in the last article.