# Area Enclosed by a Curve

In this section we learn how to calculate the area enlcosed by a curve $$y=f(x)$$ between the $$x$$-axis and two values of $$x$$, typically referred to as $$x = a$$ and $$x = b$$.
The type of area we'll know how to calculate is illustrated in the following sketch:

In each sketch, the area shaded in blue represents the area we'll soon know how to calculate.
We can see that the area corresponds to the area enclosed by the curve and the $$x$$-axis, between $$x=a$$ and $$x=b$$.

We start by learning the method, the formula, we'll then watch a tutorial and finally we'll work our way through a series of exercises.

## Formula

The area encosed by a curve between the $$x$$-axis, $$x=a$$ and $$x=b$$ can be calculated with the formula: $\text{Area} = \int_a^b \begin{vmatrix}f(x) \end{vmatrix}dx$ Where: $\begin{vmatrix} f(x) \end{vmatrix} = \begin{cases} -f(x) \quad \text{when} \quad f(x) < 0 \\ f(x) \quad \text{when} \quad f(x) \geq 0 \end{cases}$ This formula is better explained in the tutorial, below, make sure to watch it now.

## Tutorial

In the following tutorial we review the method for calculating the area enclosed by a curve and the $$x$$-axis, between two values of $$x$$.

For further explanation of the formula for the area enclosed by a curve, click on the button below.
Otherwise go straight to the exercises below.

## Explanation of the Formula

### Understanding Definite Integral Notation

To understand the formula, we've just seen, it's important to understand what is meant by the notation: $\int_a^b f(x)dx$ The following should be kept in mind:

• the integral symbol $$\int$$ is nothing more than an elogated "S", for "Sum"

• when we write $$\int_a^b$$ this means "the sum from $$a$$ to $$b$$"

• $$f(x)$$ is the height of the curve at $$x$$, meaning: if we wish to know how high the curve is at any value of $$x$$ all we have to do is calculate $$f(x)$$

• $$dx$$ is an infinitely small step across the $$x$$-axis.

• $$f(x).dx$$ is the area of an infinitely thin rectangle of height $$f(x)$$ and base width $$dx$$
Consequently, when we write $$\int_a^b f(x)dx$$ it means:

The sum of all of the infinitely thin rectangles, of height $$f(x)$$, enclosed by the curve $$y=f(x)$$ and the $$x$$-axis between $$x=a$$ and $$x = b$$.

This is illustrated in the following sketch, in which the width of the rectangles is grossly exagerated; remember: the width of each rectangle is infinitely smaller than the width of a single hair.

This tells us that, so long as the curve is above the $$x$$-axis the area enclosed by the curve $$y=f(x)$$ and the $$x$$-axis, between $$x=a$$ and $$x=b$$ is given by: $\int_a^b f(x)dx$

### The Formula for Enclosed Area

Consider the following sketch, which shows a generic curve defined as $$y=f(x)$$:

When the curve crosses the $$x$$-axis, at $$x = c$$, the values of $$f(x)$$ (equal to the $$y$$ coordinates along the curve) go from positive to negative.
A direct consequence of this is that the definite integral $$\int_c^b f(x)dx$$ is negative. The area between $$x=c$$ and $$x=b$$ therefore has a negative value.
To counteract this and to make sure all of the enclosed area is counted as positive, we consider the absolute value of $$f(x)$$, that's $$\begin{vmatrix}f(x)\end{vmatrix}$$.

This changes nothing so long as $$f(x)\geq 0$$, that's when the curve is above the $$x$$-axis, but if $$f(x)<0$$ (if the curve is below the $$x$$-axis) then the absolute value changes $$f(x)$$ to $$-f(x)$$ thereby making it postive.

The formula for the area enclosed by $$y=f(x)$$ and the $$x$$-axis between $$x=a$$ and $$x=b$$ is therefore: $\text{Area}=\int_a^b \begin{vmatrix} f(x) \end{vmatrix}dx$

## Exercise 1

Using the method just shown, calculate each of the following areas:

1. The area enclosed by $$y=x^2+1$$ and the $$x$$-axis, between $$x =0$$ and $$x = 2$$.

2. The area enclosed by $$y = x^2 - 4$$ and the $$x$$-axis, between $$x = 1$$ and $$x=3$$.

3. The area enclosed by $$y=2.sin(x)$$ and the $$x$$-axis, between $$x = 0$$ and $$x = 2\pi$$.

4. The area enclosed by $$y = \sqrt{x}+1$$ and the $$x$$-axis, between $$x = 4$$ and $$x = 9$$.

5. The area enclosed by $$y=x^2 - 4x + 3$$ and the $$x$$-axis, between $$x=1$$ and $$x = 4$$.

1. The area enclosed by $$y=x^2+1$$ and the $$x$$-axis, between $$x =0$$ and $$x = 2$$ is: $\text{Area} = \frac{14}{3}$

2. The area enclosed by $$y = x^2 - 4$$ and the $$x$$-axis, between $$x = 1$$ and $$x = 4$$ is: $\text{Area} = 4$

3. The area enclosed by $$y = 2.sin(x)$$ and the $$x$$-axis, between $$x = 0$$ and $$x = 2\pi$$ is: $\text{Area} = 8$

4. The area enclosed by $$y = \sqrt{x} + 1$$ and the $$x$$-axis, between $$x = 4$$ and $$x = 9$$ is: $\text{Area} = \frac{53}{3}$

5. The area enclosed by $$y = x^2 - 4x + 3$$ and the $$x$$-axis, between $$x = 1$$ and $$x = 4$$ is: $\text{Area} = \frac{8}{3}$

## Solution With Working

1. #### Question 1

To find the area we follow our three steps:
• Step 1: sketch the curve and highlight the area that needs to be calculated.

The sketch of $$y=x^2+1$$ and the area that needs to be calculated is shown here:

We've shaded the area that needs to be calculated in blue, and written "$$+ \ \text{ve}$$" to highlight the fact that the area there will be positive.

• Step 2: find any intervals over which the curve is below the $$x$$-axis and state the formula for the area.

Looking at our sketch, above, it's clear that the entire length of the curve is above the $$x$$-axis.
This means that $$x^2+1 \geq 0$$, so we don't need to use the absolute value.

The formula for the shaded area is therefore: $\text{Area} = \int_0^2 \begin{pmatrix} x^2+1 \end{pmatrix}dx$

• Step 3: calculate the area. \begin{aligned} \text{Area} &= \int_0^2 \begin{pmatrix} x^2+1 \end{pmatrix}dx\\ & = \begin{bmatrix} \frac{x^3}{3} + x \end{bmatrix}_0^2 \\ & = \begin{bmatrix} \frac{2^3}{3} + 2 \end{bmatrix} - \begin{bmatrix} \frac{0^3}{3} + 0 \end{bmatrix} \\ & = \begin{bmatrix} \frac{8}{3} + 2 \end{bmatrix} - \begin{bmatrix} 0 + 0 \end{bmatrix} \\ & = \begin{bmatrix} \frac{8}{3} + \frac{6}{3} \end{bmatrix} - \begin{bmatrix} 0 \end{bmatrix} \\ & = \frac{14}{3} - 0 \\ \text{Area} & = \frac{14}{3} \quad \text{(approx. 4.67)} \end{aligned}
Finally, we state that the area enclosed by the curve and the $$x$$-axis, between $$x=0$$ and $$x = 2$$ is equal to $$\frac{14}{3}$$ units of area.

2. #### Question 2

To calculate the area, we follow our three step method:
• Step 1: make a sketch of a curve, $$y=x^2-4$$ to picture the area that needs ot be calculated.

The sketch is done below, we've shaded the area that needs to be calculated in blue and highlighted the fact that the "area is positive" by writing a "$$+$$ ve" inside the area:

• Step 2: find any intervals over which the curve is below the $$x$$-axis and state the formula to use to calculate the area.

Looking at this sketch we make a note of the following important facts:

• for $$1\leq x \leq 2$$, the curve is below the $$x$$-axis. This means that the definite integral $$\int_1^2 \begin{pmatrix} x^2 - 4 \end{pmatrix} dx$$ will be negative, we'll therefore have to use the absolute value of the function; we can think of this as a "negative area".

• for $$2 \leq x \leq 3$$, the curve is above the $$x$$-axis. The definite integral $$\int_2^3 \begin{pmatrix} x^2-4 \end{pmatrix} dx$$ will be positive, so no absolute value will be needed.

Keeping those two facts in mind, the formula for calculating the area is: \begin{aligned} \text{Area} &= \int_1^3 \begin{vmatrix} x^2 - 4 \end{vmatrix} dx \\ & = \int_1^2 -\begin{pmatrix} x^2 - 4 \end{pmatrix} dx + \int_2^3 \begin{pmatrix} x^2 - 4 \end{pmatrix}dx \\ \text{Area} & = -\int_1^2 \begin{pmatrix} x^2 - 4 \end{pmatrix} dx + \int_2^3 \begin{pmatrix} x^2 - 4 \end{pmatrix} dx \end{aligned} Make a Note: it's worth pointing out that "all" that taking the absolute value of the function does is: multiply the integral by $$-1$$ as soon as the curve is beneath the $$x$$-axis.

• Step 3: calculate the area. \begin{aligned} \text{Area} &= -\int_1^2 \begin{pmatrix} x^2-4 \end{pmatrix} dx + \int_2^3 \begin{pmatrix} x^2 - 4 \end{pmatrix} dx \\ & = - \begin{bmatrix} \frac{x^3}{3} - 4x \end{bmatrix}_1^2 + \begin{bmatrix} \frac{x^3}{3} - 4x \end{bmatrix}_2^3 \\ & = - \begin{bmatrix} \begin{pmatrix} \frac{2^3}{3} - 4 \times 2 \end{pmatrix} - \begin{pmatrix} \frac{1^3}{3} - 4 \times 1 \end{pmatrix} \end{bmatrix} + \begin{bmatrix} \begin{pmatrix} \frac{3^3}{3} - 4 \times 3 \end{pmatrix} - \begin{pmatrix} \frac{2^3}{3} - 4 \times 2 \end{pmatrix} \end{bmatrix} \\ & = - \begin{bmatrix} \begin{pmatrix} \frac{8}{3} - 8 \end{pmatrix} - \begin{pmatrix} \frac{1}{3} - 4 \end{pmatrix} \end{bmatrix} + \begin{bmatrix} \begin{pmatrix} \frac{27}{3} - 12 \end{pmatrix} - \begin{pmatrix} \frac{8}{3} - 8 \end{pmatrix} \end{bmatrix} \\ & = - \begin{bmatrix} \begin{pmatrix} -\frac{16}{3} \end{pmatrix} - \begin{pmatrix} -\frac{11}{3} \end{pmatrix} \end{bmatrix} + \begin{bmatrix} \begin{pmatrix} -3 \end{pmatrix} - \begin{pmatrix} -\frac{16}{3} \end{pmatrix} \end{bmatrix} \\ & = - \begin{bmatrix} -\frac{5}{3} \end{bmatrix} + \begin{bmatrix} \frac{7}{3} \end{bmatrix} \\ & = \frac{5}{3} + \frac{7}{3} \\ & = \frac{12}{3} \\ \text{Area} & = 4 \quad \text{(units of area)} \end{aligned}
Finally, we state that the area enclosed by the curve and the $$x$$-axis, between $$x=1$$ and $$x = 3$$ is equal to $$4$$ units of area.

3. #### Question 3

To calculate the area, we follow our three step method:
• Step 1: we make a sketch of the curve, $$y=2.sin(x)$$, to picture the area that needs ot be calculated.

We've shaded the area that needs to be calculated. The blue area corresponds to the area above the $$x$$-axis, it corresponds to "positive area". The pink area corresponds to the area below the $$x$$-axis, it corresponds to "negative area", for which we'll need to use the absolute value.

• Step 2: find any intervals over which the curve is beneath the $$x$$-axis and state the formula that needs to be used to calculate the area.

Looking at our sketch, above, we see that:
• for $$0 \leq x < \pi$$, the curve is above the $$x$$-axis. We won't need to use the absolute value over this interval.

• for $$\pi \leq x < 2 \pi$$, the curve is below the $$x$$-axis. We'll need to use the absolute value over this interval.
We can now state that the enclosed area is equal to: \begin{aligned} \text{Area} & = \int_0^{2\pi} \begin{vmatrix} 2.sin(x) \end{vmatrix}dx \\ & = \int_0^{\pi} 2.sin(x)dx + (-1)\times \int_{\pi}^{2\pi} 2.sin(x) dx \\ & = \int_0^{\pi} 2.sin(x)dx -\int_{\pi}^{2\pi} 2.sin(x) dx \\ \text{Area} & = 2\int_0^{\pi} sin(x)dx -2 \int_{\pi}^{2\pi} sin(x) dx \end{aligned} Remember: "all" that taking the absolute value of the function does is: multiply the integral by $$-1$$ as soon as the curve is beneath the $$x$$-axis.

• Step 3: calculate the area: \begin{aligned} \text{Area} & = 2\int_0^{\pi} sin(x)dx -2 \int_{\pi}^{2\pi} sin(x) dx \\ & = 2 \begin{bmatrix} - cos(x) \end{bmatrix}_0^{\pi} - 2 \begin{bmatrix} - cos(x) \end{bmatrix}_{\pi}^{2 \pi} \\ & = 2 \begin{bmatrix} -cos(\pi ) - \begin{pmatrix} -cos(0) \end{pmatrix} \end{bmatrix} - 2 \begin{bmatrix} -cos(2 \pi ) - \begin{pmatrix} -cos(\pi) \end{pmatrix} \end{bmatrix} \\ & = 2 \begin{bmatrix} - (-1) - (-1) \end{bmatrix} - 2 \begin{bmatrix} - 1 - \begin{pmatrix} - (-1) \end{pmatrix}\end{bmatrix}\\ & = 2 \begin{bmatrix} 1 + 1 \end{bmatrix} - 2 \begin{bmatrix} -1 - 1 \end{bmatrix} \\ & = 2\begin{bmatrix} 2 \end{bmatrix} - 2\begin{bmatrix} - 2 \end{bmatrix} \\ & = 2\times 2 - 2 \times (-2) \\ & = 4 - (-4) \\ & = 4+4 \\ \text{Area} & = 8 \quad \text{(units of area)} \end{aligned}
Finally, we state that the area enclosed by the curve and the $$x$$-axis, between $$x=0$$ and $$x = 2 \pi$$ is equal to $$8$$ units of area.

4. #### Question 4

• Step 1: sketch the curve and the area that needs to be calculated.

This is done here:
The area that needs to be calculated is shaded in blue, with "$$+$$ ve" written inside it to highlight the fact that it is above the $$x$$-axis.

• Step 2: find any intervals over which the curve is below the $$x$$-axis and state the formula that will be used for the area.

The curve's entire length is above the $$x$$-axis, so $$\sqrt{x}+1 \geq 0$$. The sbolute value won't be needed.

The area enclosed by the curve and the $$x$$-axis, between $$4$$ and $$9$$ is therefore given by the formula: $\text{Area} = \int_4^9 \begin{pmatrix} \sqrt{x}+1 \end{pmatrix}dx$

• Step 3: calculate the area.

\begin{aligned} \text{Area} & = \int_4^9 \begin{pmatrix} \sqrt{x}+1 \end{pmatrix}dx \\ & = \int_4^9 \begin{pmatrix} x^{\frac{1}{2}}+1 \end{pmatrix}dx \\ & = \begin{bmatrix} \frac{2}{3}x^{\frac{3}{2}} + x \end{bmatrix}_4^9 \\ & = \begin{bmatrix} \frac{2}{3}\times 9^{\frac{3}{2}} + 9 \end{bmatrix} - \begin{bmatrix} \frac{2}{3}\times 4^{\frac{3}{2}} + 4 \end{bmatrix} \\ & = \begin{bmatrix} \frac{2}{3}\times \begin{pmatrix} 9^{\frac{1}{2}} \end{pmatrix}^3 + 9 \end{bmatrix} - \begin{bmatrix} \frac{2}{3}\times \begin{pmatrix} 4^{\frac{1}{2}} \end{pmatrix}^3 + 4 \end{bmatrix} \\ & = \begin{bmatrix} \frac{2}{3}\times \begin{pmatrix} 3 \end{pmatrix}^3 + 9 \end{bmatrix} - \begin{bmatrix} \frac{2}{3}\times \begin{pmatrix} 2 \end{pmatrix}^3 + 4 \end{bmatrix} \\ & = \begin{bmatrix} \frac{2}{3} \times 27 + 9 \end{bmatrix} - \begin{bmatrix} \frac{2}{3} \times 8 + 4 \end{bmatrix} \\ & = \begin{bmatrix} \frac{54}{3} + 9 \end{bmatrix} - \begin{bmatrix} \frac{16}{3} + 4 \end{bmatrix} \\ & = \begin{bmatrix} 27 \end{bmatrix} - \begin{bmatrix} \frac{28}{3} \end{bmatrix} \\ & = \frac{81}{3} - \frac{28}{3} \\ \text{Area} & = \frac{53}{3} \end{aligned}
Finally, we state that the area enclosed by the curve and the $$x$$-axis, between $$x=4$$ and $$x = 9$$ is equal to $$\frac{53}{3}$$ units of area.

5. #### Question 5

We use our three-step method:
• Step 1: sketch the curve $$y = x^2 - 4x +3$$ and highlight the area that needs to be calculated.

This is sketched here:
The area beneath the $$x$$-axis is shaded in pink with "$$-$$ ve" written inside it and the area above the $$x$$-axis is shaded in blue with "$$+$$ ve" written.

• Step 2: find any intervals over which the curve is below the $$x$$-axis.

Looking at our sketch, we see that:
• for $$1\leq x < 3$$, the curve is below the $$x$$-axis. The area there is therefore nagative and we'll have to use the absolute value.

• for $$3 \leq x \leq 4$$, the curve is above the $$x$$-axis. The area there is therefore positive and we won't need to use the absolute value.
The formula for calculating the area enclosed by the curve will therefore be: \begin{aligned} \text{Area} & = \int_1^4 \begin{vmatrix} x^2 - 4x + 3 \end{vmatrix} dx \\ & = \int_1^3 \begin{vmatrix} x^2 - 4x + 3 \end{vmatrix} dx + \int_3^4 \begin{vmatrix} x^2 - 4x + 3 \end{vmatrix} dx \\ \text{Area}& = (-1)\times \int_1^3 \begin{pmatrix} x^2 - 4x + 3 \end{pmatrix} dx + \int_3^4 \begin{pmatrix} x^2 - 4x + 3 \end{pmatrix} dx \\ \end{aligned}

• Step 3: calculate the area.

\begin{aligned} \text{Area}& = (-1)\times \int_1^3 \begin{pmatrix} x^2 - 4x + 3 \end{pmatrix} dx + \int_3^4 \begin{pmatrix} x^2 - 4x + 3 \end{pmatrix} dx \\ & = (-1)\times \begin{bmatrix} \frac{x^3}{3} - 2x^2 + 3x \end{bmatrix}_1^3 + \begin{bmatrix} \frac{x^3}{3} - 2x^2 + 3x \end{bmatrix}_3^4 \\ & = (-1) \times \begin{bmatrix} \begin{pmatrix} \frac{3^3}{3} - 2\times 3^2 + 3 \times 3 \end{pmatrix} - \begin{pmatrix} \frac{1^3}{3} - 2\times 1^2 + 3 \times 1 \end{pmatrix} \end{bmatrix} + \begin{bmatrix} \begin{pmatrix} \frac{4^3}{3} - 2\times 4^2 + 3 \times 4 \end{pmatrix} - \begin{pmatrix} \frac{3^3}{3} - 2\times 3^2 + 3 \times 3 \end{pmatrix} \end{bmatrix} \\ & = (-1)\times \begin{bmatrix} \begin{pmatrix} \frac{27}{3} - 2\times 9 + 9 \end{pmatrix} - \begin{pmatrix} \frac{1}{3} - 2 + 3 \end{pmatrix} \end{bmatrix} + \begin{bmatrix} \begin{pmatrix} \frac{64}{3} - 2 \times 16 + 12 \end{pmatrix} - \begin{pmatrix} \frac{27}{3} - 2 \times 9 + 9 \end{pmatrix} \end{bmatrix} \\ & = (-1) \times \begin{bmatrix} 0 - \frac{4}{3} \end{bmatrix} + \begin{bmatrix} \frac{4}{3} - 0 \end{bmatrix} \\ & = (-1) \times \begin{bmatrix} - \frac{4}{3} \end{bmatrix} + \begin{bmatrix} \frac{4}{3} \end{bmatrix} \\ & = \frac{4}{3} + \frac{4}{3} \\ \text{Area} & = \frac{8}{3} \quad \text{(approx. 2.67)} \end{aligned}
Finally, we state that the area enclosed by the curve and the $$x$$-axis, between $$x=1$$ and $$x = 4$$ is equal to $$\frac{8}{3}$$ units of area.

## Calculating Areas with a Calculator

On top of knowing how to calculate by hand, it is very important that we know how to calculate the area enclosed by a curve using a Graphical Calculator.
Indeed, this is particularly important, as some integrals cannot easily be found by hand.

Important: when calculating areas under curves with a calculator make sure to always use the absolute value of the function.

The method for doing this is shown here, for the TI NSPire CX.

## Exercise 2

Using the method just shown and rounding your answers to $$2$$ d.p, calculate each of the following areas with your calculator:

1. The area enclosed by $$y = ln^2(x).sin^3(x-2)dx$$ and the $$x$$-axis, between $$x = 2$$ and $$x = 5$$.

2. The area enclosed by $$y =\frac{sin^{-1}(x^2+2)}{e^{2x+1}}dx$$ and the $$x$$-axis, between $$x = 1$$ and $$x = 2$$.

3. The area enclosed by $$y = x^3.tan^2(2x+1)\sqrt{x+4}dx$$ and the $$x$$-axis, between $$x = 0$$ and $$x = 5$$.

4. The area enclosed by $$y = 2^{-x}\begin{pmatrix}x^3-x^2+x+3 \end{pmatrix}dx$$ and the $$x$$-axis, between $$x = 0$$ and $$x = 3$$.

## Solution Without Working

1. $$\text{Area} = 2.15$$