A Brief Introduction to Calculus: Part 2.2 - Area Under the Curve

Math
Calculus
Further Math

Published on Fri Aug 05 2022. Views

An introduction to basic concepts in Calculus; mostly based on the IGCSE Further Math textbook.

Introduction

This series would primarily focus on the three fundamental areas of calculus - derivatives, limits, and integration, looking briefly at some of the different concepts to get started with calculus.; thus no prior knowledge of calculus is needed.

This is the second chapter of the second part of the series, and it is about finding the area under the curve.

Further content on this topic will be published in separate articles for performance reasons.

So now, let's sit back, relax, and have some fun!

Quick Access

Using Integration to Find the Area

Firstly, let's see how we can use definite integrals to calculate the exact area of a region bounded by the $x$ -axis, and the lines $x = a$ and $x = b$ .

The method for Riemann sum, which will be discussed later, is an approximation of the integral, because the value calculated from the integral is the limit of the Riemann sum, meaning that the approximation will always converge(approach) on the value of the integral.

The Basic Scenario

When the entirety of the region required is on one side of the $x$ -axis, we can find the area between the curve and the $x$ -axis, bounded by the limits $[a, b]$ using the definite integral:

\begin{align*} \int_{a}^{b} f(x)\, dx \end{align*}

Examples

The function of the curve is given by $f(x) = -x^2 + 10$ , find the exact area of the shaded region.

\begin{align*} a &= -2 \\ b &= 2 \\ \\ A &= \int_{-2}^{2} \left[-x^2 + 10\right]\, dx \\ &= \left[-\frac{1}{3}x^3 + 10x\right]_{-2}^{2} \\ &= 34\frac{2}{3} \end{align*}

The function of the curve is given by $f(x) = \left(x-4\right)\left(x+4\right)$ , find the exact area of the shaded region.

\begin{align*} I &= \int_{0}^{4} (x-4)(x+4)\, dx \\ &= \int_{0}^{4} \left(x^2 - 16\right)\, dx \\ &= \left[\frac{x^3}{3} - 16x\right]_{0}^{4} \\ &= -\frac{128}{3} \end{align*}

Now, because (in this context) area is a scalar quantity, we should take the absolute value of the integral.

\begin{align*} \therefore A &= |-\frac{128}{3}| \\ &= 42\frac{2}{3} \end{align*}

When the Region Is on Both Sides of the X-Axis

Consider the following case:

The function of the curve is given by $f(x) = -3\sin(x)$ . In the range of values we are looking at, the curve intercepts the x-axis at $x \in \{-\pi, 0, \pi\}$ .

If you set $a = -\pi$ and $b = \pi$ and use integration to find the area:

\begin{align*} A &= \int_{-\pi}^{\pi} \left[-3\sin(x)\right]\, dx \\ &= \Big[3\cos(x)\Big]_{-\pi}^{\pi} \\ &= -3 - (-3) = 0 \end{align*}

However, the area is clearly not 0. So how should we go about finding the area for regions like this?

To compute the correct area, split the region into different parts by their direction relative to the x-axis. In this case, calculate the area above the x-axis and add the area below the x-axis to obtain the final answer.

Note that when adding the respective parts, take the absolute value of the integral where necessary to avoid the subtraction of areas due to a negative result from an integral.

\begin{align*} A &= \int_{-\pi}^{0} -3\sin(x)\, dx+ \left| \int_{0}^{\pi} -3\sin(x)\, dx\right| \\ &= 3 + \mid-3\mid \\ &= 6 \end{align*}

Example

The function of the curve below is given by $f(x) = x(x - 1)(x + 3)$ .

Find the area of the finite region bounded by the x-axis and the curve.

\textnormal{When }y = 0\textnormal{, }x \in \{-3, 0, 1\}

\begin{align*} &x(x - 1)(x + 3) \\ =& x[x^2 + 2x - 3] \\ =& x^3 + 2x^2 - 3x \\ \end{align*}

\begin{align*} A &= \int_{-3}^{0} f(x)\, dx + \left|\int_{0}^{1} f(x)\, dx\right| \\ \\ F(x) &= \frac{x^4}{4} + \frac{2}{3}x^3 - \frac{3}{2}x^2 \\ A &= F(0) - F(-3) + \left|F(1) - F(0)\right| \\ &= 11\frac{5}{6} \end{align*}

Riemann Sum

The Riemann sum is a strategy for estimating the area under a curve by dividing the area into $n$ rectangular strips, where each rectangle has a defined width, namely $\Delta x$ , and its height is defined by the $y$ value calculated using each x value $f(x_i)$ .

There are three types of Riemann sums:

Right endpoint sum
Left endpoint sum
Midpoint rule

We will discuss the differences later.

The formula is as follows, however, for the different types of Riemann sum, the formulas differ a little.

\begin{align*} A \approx \sum_{i=1}^{n}\, \Delta x f(x_i) \end{align*}

Let's break this down step by step.

$n$ is the number of strips the area is divided into; generally the higher $n$ is, the closer the approximation
$\Delta x$ is the width of each rectangle, and it's given by $\frac{b - a}{n}$ , where $b$ and $a$ are the limits on the x-axis
$f(x)$ is the function of the curve, and for a given $x_i$ , it returns the height of the rectangle
$x_i$ is the ith value in the arithmetic sequence, given by $x_i = x_1 + (i - 1)\Delta x$ , which is based on $U_i = U_1 + (n-1)d$
$x_1 = a$ , which means that the sequence starts at the lower bound, hence $x_i = a + (i - 1)\Delta x$

Furthermore, let $R(n)$ be the function for the left endpoint Riemann approximation:

\begin{align*} R(n) = \sum_{i=1}^{n}\, \Delta x f(x_i) \end{align*}

As $n$ , the number of subintervals is increased, the blank gap reduces, therefore increasing the resulting approximated area:

However, the value of $R(n)$ will never exceed a certain value no matter how large $n$ becomes, and that limit is the exact area of the curve.

By definition, the integral used to find the area for the same region is exactly the limit of $R(n)$ as $n \rightarrow \infty$ :

\begin{align*} \int_{a}^{b} f(x)\, dx = \lim_{n \to \infty} R(n) \end{align*}

Left Endpoint, Right Endpoint, and Midpoint

For a left endpoint sum, each rectangle's top-left corner touches the curve, and thus the approximation is always below the actual area.

In comparison, for a right endpoint sum, each rectangle's top-right corner touches the curve, therefore the approximation will always be above the actual area.

Note that, in this case, the height of the rectangles is $f(x)$ for which $x$ is shifted by $\Delta x$ to the right, thus:

\begin{align*} A \approx \sum_{i=1}^{n}\, \Delta x f(x_i + \Delta x) \end{align*}

Finally, for a midpoint rule, the curve meets each strip at the midpoint of the length of the rectangle.

Likewise, for the midpoint rule, the height of the rectangles is $f(x)$ for which $x$ is shifted by $\frac{1}{2}\Delta x$ to the right, thus:

\begin{align*} A \approx \sum_{i=1}^{n}\, \Delta x f(x_i + \frac{1}{2}\Delta x) \end{align*}

Example

Calculate the area under the curve $f(x)=x^{2} + 1$ bounded by $x = 0$ and $x = 3$ . Use the midpoint rule.

A \approx \sum_{i=1}^{n}\, \Delta x f(x_i + \frac{1}{2}\Delta x)\\ a = 0 \\ b = 3

\begin{align*} \textnormal{Let } n &= 18 \\ \\ \Delta x &= \frac{3 - 0}{18} = \frac{1}{6} \\ A &\approx \sum_{i=1}^{18}\, \frac{1}{6} \times f(a + \frac{1}{6}(i - 1) + \frac{1}{2} \times \frac{1}{6}) \\ &\because a = 0 \\ A &\approx \sum_{i=1}^{18}\, \frac{1}{6} \times f(\frac{1}{6}(i - 1) + \frac{1}{12}) \\ A &\approx \frac{1727}{144} \\ &\approx 11\frac{143}{144} \end{align*}

You will realize that the value is very close to what the definite integral would give:

\begin{align*} \int_{0}^{3} f(x)\, dx = 12 \end{align*}

Conclusion

This blog introduced a few basic concepts of definite and indefinite integrals, and more content will be covered in future blogs of this part of the series.

Thanks for reading!