Common Applications of Integrals

These notes provide an overview of places where integrals frequently show up.

Weighted Average
Probability
Numerical Integration

Weighted Average

The idea of a weighted average is that we take the integral of some function from ${a}$ to ${b,}$ and multiply it by some weight, and divide it by the total weighting:

\dfrac {\displaystyle \integral{a}{b} f(x) \cdot w(x)~dx} {\displaystyle \integral{a}{b} w(x) dx}

This definition isn't that unusual. Think about the average we computed in grade school. We have some list of values:

\ix{v_1, v_2, v_3, v_4, \ldots, v_n}

and we divide the sum of all these values by the number of terms:

\dfrac {v_1 + v_2 + v_3 + v_4 + \ldots + v_n} {n}

This is the discrete version of an average — what we know as the arithmetic mean. The continuous version of an average is the definition above: a weighted average. Of note, many statistics textbooks often define the weighted average as:

\overline{x}_{wtd} = \dfrac {\dsum{i=1}{n} w_i x_i} {\dsum{i=1}{n} w_i}

The notion of a weighted average brings us to one of the most important applications of integration in the modern era: probability.

Probability

Suppose we had the following graph:

if we picked a point "at random" in the region ${0 \ltn y \ltn 1-x^2,}$ what is the probability that ${x \gtn \frac{1}{2}?}$ Casting this question in probability terms, this question asks for the probability:

\P \ar{x \gtn \frac{1}{2}}

From the perspective of calculus, a probability is simply a special kind of ratio. Specifically, a ratio of the part of something to the whole of something:

\dfrac{\text{part}}{\text{whole}}

The part that we're interested in:

This is a straightforward integration problem. That area is:

\dfrac{\displaystyle \int_{\frac{1}{2}}^{1} 1-x^2 ~dx} {\displaystyle \int_{-1}^{1} 1-x^2 ~dx}

As we saw with the weighted average, the weighting factor here is:

w(x) = 1-x^2

And just to be more explicit, our starting point is ${-1,}$ and our ending point is ${b = 1.}$ Working out the computation, we get:

\P \ar{x \gtn \frac{1}{2}} = \dfrac{5}{32}

We can generalize this computation with the following formula:

\P(x_1 \ltn x \ltn x_2) = \dfrac {\text{part}} {\text{whole}} = \dfrac {\displaystyle \int_{x_1}^{x_2} w(x) ~ dx} {\displaystyle \int_{a}^{b} w(x) ~ dx}

Let's consider an example taken from the real world. Suppose we're on a manhunt for a wanted criminal. We've narrowed down their possible location to the following:

The question: What are the odds that the crook is somewhere in the blue region? To simplify our computation, we make the following assumption:

The number of hits is proportional to ${ce^{-r^2},}$ where ${c}$ is a constant, ${e}$ is Euler's number, and ${r}$ is the radius.

This assumption effectively states that the number of hits is a normal distribution in terms of a given a radius. To understand what this assumption implies, let's take a look at the plot of the function ${f = e^{-r^2}.}$

Suppose that, instead of a manhunt, we were throwing darts. When we make the assumption above, we're stating that, if we threw millions of darts, more of them would land closer to the middle (the top of the bell curve), and fewer of them would land towards the lower and upper ends.

Question: Why specifically do we use the function ${f = e^{-r^2}?}$ Why not some other function? This is probability question rather than a calculus question, so we won't answer it in detail (it's akin to asking why we would use one kinematic equation over another given a physics problem). It's simply the most accurate model for our purposes. When the Germans bombed London during the Second World War, the British miliary found that among all the possible functions they could use to predict where the bombs might land, the function ${f = e^{-r^2}}$ proved to be the most accurate.