## Introduction

Look at the following graph of the function $$f(x) = \sqrt{4-x^2}, \quad -2 < x < 2$$ and the region highlighted in orange. Can you compute its area?

As you may have noticed, the highlighted area is one quarter of a circle. Since the area of a full circle is $$\pi r^2$$ where $$r$$ is the circle’s radius and the radius $$r=2$$, the area equals

$$A = \frac{1}{4} \pi r^2 = \frac{1}{4} \pi (2)^2 = \pi.$$

But now suppose we would like to find the area under the graph of $$f(x)$$, in between $$x=0$$ and some other given value of $$x > 0$$. Hover over the following graph to see such a region being highlighted.

Is there any way we can define and compute the area of such a region?

## What is the integral?

This leads to what we call the Riemann integral. Consider first a nonnegative function $$f(x)\geq 0$$. To find the area below the graph in the interval $$a < x < b,$$ first subdivide the interval $$[a, b]$$ into a large number of smaller intervals: $$a=x_0 < x_1 < x_2 < \ldots < x_{n-1} < x_n = b.$$ This just means that we have picked numbers $$x_i$$ lying in between $$a$$ and $$b$$. In a second step, we can then go ahead and pick points inside the small intervals $$[x_i, x_{i+1}]$$. We will call these number $$y_i$$ and they have the property $$x_{i} \leq y_i \leq x_{i+1}.$$ With all of that preparation, we can define a typical Riemann sum:

$$\sum_{i=0}^{n-1} f(y_i) \cdot (x_{i+1}-x_i)$$

This looks complicated, but is in effect the summation of the areas of slim rectangles. Use the following simulation to visualize how the Riemann sum works: first use the slider to choose a step size $$h$$. For simplicity, we have fixed the size of all the small intervals to be $$h > 0$$, that is $$x_{i+1}-x_{i} = h.$$

Hover over the image and play around with the stepsize. The buttons allow you to adjust the intermediate points of the Riemann sum. Can you compute the area of the quarter circle with radius $$2$$, and do you get the expected answer?

## How are integrals computed?

We can define the integral of any continuous function $$f(x)$$ that is defined on a closed interval $$[a, b]$$ by a limit process. For this, we imagine making $$h$$ successively smaller, eventually approaching $$0$$. Although the Riemann sums differ in value for different $$h$$, the process can be shown to converge: the Riemann sums approach a number for which we use the special symbol $$\int_a^b f(x)~dx$$:

$$\int_a^b f(x) ~dx = \lim_{h\to 0}\sum_{i=0}^{n-1} f(y_i) \cdot (x_{i+1}-x_i)$$

The number $$\int_a^b f(x)~dx$$ can be interpreted as area under the graph of the function.

What does it mean if $$\int_a^b f(x)~dx$$ is negative?

A few basic rules are:

$$\int_a^b (f(x) + g(x))~dx = \int_a^b f(x)~dx + \int_a^b g(x)~dx$$ $$\int_a^b k f(x)~dx = k \int_a^b f(x)~dx\quad\text{for any number k}$$ $$\int_a^b f(x)~dx = \int_a^c f(x)~dx + \int_c^b g(x)~dx\quad \text{if a < c < b}$$

Another way to compute integrals is to use the Fundamental Theorem of Calculus. We will come back to this in a later post.

## Why are integrals useful?

Some applications of integrals are:

• Areas, volumes and lengths of curves in any number of dimension
• Averages over time: $$A = \frac{1}{T}\int_0^T f(t)~dt$$ is a time average of $$f(t)$$ from $$0$$ to $$T$$
• In physics: $$\int \vec{F}\cdot d\vec{s}$$ with $$\vec{F}$$ the force acting on an object is the work
• Integration is the inverse operation of taking derivatives; more about that in a future post!
• Approximating complicated sums, such as $$\sum_{n=1}^\infty \frac{1}{n^2}$$ and evaluating complicated limits, such as $$\lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^n \sin\left(\frac{k\pi}{n}\right)$$