In this article, we will discuss how to calculate the area between two functions. We will specifically concentrate on how to calculate the area between a curve and a straight line, and the area between two curves.

Area Between Two Functions

The area between two functions is equal to the area of the function located above minus the area of the function that lies below. Mathematically, we can denote this area like this:

A = \int^b_a [g(x) - f(x)] dx

 

The best Maths tutors available
1st lesson free!
Intasar
4.9
4.9 (26 reviews)
Intasar
£36
/h
1st lesson free!
Matthew
5
5 (17 reviews)
Matthew
£25
/h
1st lesson free!
Paolo
4.9
4.9 (13 reviews)
Paolo
£25
/h
1st lesson free!
Dr. Kritaphat
4.9
4.9 (7 reviews)
Dr. Kritaphat
£49
/h
1st lesson free!
Ayush
5
5 (28 reviews)
Ayush
£60
/h
1st lesson free!
Petar
4.9
4.9 (9 reviews)
Petar
£27
/h
1st lesson free!
Farooq
5
5 (14 reviews)
Farooq
£40
/h
1st lesson free!
Tom
5
5 (9 reviews)
Tom
£22
/h
1st lesson free!
Intasar
4.9
4.9 (26 reviews)
Intasar
£36
/h
1st lesson free!
Matthew
5
5 (17 reviews)
Matthew
£25
/h
1st lesson free!
Paolo
4.9
4.9 (13 reviews)
Paolo
£25
/h
1st lesson free!
Dr. Kritaphat
4.9
4.9 (7 reviews)
Dr. Kritaphat
£49
/h
1st lesson free!
Ayush
5
5 (28 reviews)
Ayush
£60
/h
1st lesson free!
Petar
4.9
4.9 (9 reviews)
Petar
£27
/h
1st lesson free!
Farooq
5
5 (14 reviews)
Farooq
£40
/h
1st lesson free!
Tom
5
5 (9 reviews)
Tom
£22
/h
First Lesson Free>

Area Between a Curve and a Straight Line

Now, let us understand how to compute the area between a curve and a straight line through the following examples

 

Example 1

Find the area of the space bounded by the parabola y = x^2 + 2 and the straight line that passes through the points A(−1, 0) and B(1, 4).

Solution

Step 1 - Find the equation of a straight line

In this step, we will calculate the equation of a straight line passing through two points A and B. To do so, first, we should calculate the slope of the line that passes through the points A(-1, 0) and B(1, 4). To calculate the slope, we will use the following formula:

m = \frac{y_2 - y_1} {x_2 - x_1}

Substitute the values of the points A and B in the above formula:

m = \frac{4 - 0} {1 - (-1)} = 2

Now, substitute this slope in the point intercept equation below:

y - y_1 = m(x - x_1)

y - 0 = 2x + 2

y = 2x + 2

Hence, the equation of the straight line is y = 2x + 2.

Step 2 - Sketch the graph 

In this step, we will sketch the graph of the function y = x^2 + 2 and line y = 2x + 2 like this:

Example 1 - Graph of the straight line and the curve

Step 3 - Calculate the boundaries

The points at which the line intersect the parabolas will be the boundaries or limits of the function. As we can see from the above graph, that  the line intersects the parabola at x_1 = 0 and x_2 = 2. Hence, these are limits of the function.

Step 4 - Calculate the definite integral

To calculate the definite integral, first, use the information from the previous steps to write the functions in the following form:

A = \int^b_a [g(x) - f(x)] dx

A = \int^2_0 (2x + 2 - x^2 - 2)dx

A = \int^2_0 (2x - x^2)dx

Rewrite the function \int^{4}_{-1} (4 f(x) - 3 g(x)) dx by using the sum/difference rule of definite integrals like this:

\int^{2}_{0} 2x dx - \int^{2}_{0} x^2 dx

To compute  the definite integral, we will first find the antiderivative of the function. The antiderivative of the function is x^2 - \frac{x^3}{3}

Now, use the fundamental theorem of calculus:

\int _{a} ^ {b} f(x) = [F(x)]_{a} ^{b} = F (b) - F(a)

Substitute 2 and 0 in the antiderivative of the function like this:

A = (4 - \frac{8}{3}) - (0) = \frac{4}{3} u^2

 

Example 2

Calculate the area of the figure bounded by the function y = x^2 and the lines y = x, at x = 0 and x = 2.

Solution

Step 1 - Sketch the graph

In this example, we are already given the equation of the line y = x. Hence, we don't need to calculate it. We will simply start by sketching the graph of the functions y = x^2, and y = x.

Example 1 - Area between a curve and a straight line

You can see in the above graph that from x = 0 to x = 1, the straight line is above the parabola, and from x = 1 to x = 2, the straight line is below the parabola. Hence, we will compute the areas using these limits above and below the parabola separately.

 

Step 2 - Calculate the boundaries

The boundaries or limits of the graph are already given in this example which are 0 and 1.

Step 3 - Calculate the definite integral

To calculate the definite integral, first, use the information from the previous steps to write the functions in the following form:

A = \int^b_a [g(x) - f(x)] dx

Area where straight line is above the parabola:

A = \int^1_0 (x - x^2)dx

Find the antiderivative of the function. The antiderivative of the function is \frac{x^2}{2} - \frac{x^3}{3}

Use the fundamental theorem of calculus:

\int _{a} ^ {b} f(x) = [F(x)]_{a} ^{b} = F (b) - F(a)

Substitute 1 and 0 in the antiderivative of the function like this:

A = \frac{1}{2} - \frac{1}{3} - 0

A = \frac{1}{6} u^2

Area where straight line is below the parabola:

A = \int^2_1 ( x^2 - x)dx

Find the antiderivative of the function. The antiderivative of the function is \frac{x^3}{3} - \frac{x^2}{2}

Use the fundamental theorem of calculus:

\int _{a} ^ {b} f(x) = [F(x)]_{a} ^{b} = F (b) - F(a)

Substitute 2 and 1 in the antiderivative of the function like this:

A =  \frac{2^3}{3} -\frac{2^2}{2} - \frac{1^3}{3} - \frac{1^2}{2}

A = \frac{5}{6} u^2

 

In the next section, we will see how to calculate the area between two curves given their equations.

 

Area Between Two Curves

The following examples will let you understand how to calculate the area between two curves.

Example 1

Find the area bounded by the graphs of the functions y^2 = 4x and y = x^2

Solution

Step 1 - Sketch the graph

Example 1 - Area between two curves

Step 2 - Find the boundaries

To determine where the graphs of two curves intersect each other, we will equate the equations of two curves:

4x = (x^2)^2

x^4 - 4x = 0

x (x^3 - 4) = 0

x = 0 or x = \sqrt[3]{4}

Hence, the boundaries are \sqrt[3]{4} and 0.

 

Step 3 - Calculate the definite integral

To calculate the definite integral, first, use the information from the previous steps to write the functions in the following form:

A = \int^b_a [g(x) - f(x)] dx

A = \int^{\sqrt[3]{4}}_0 (2 \sqrt{x} - x^2)dx

Find the antiderivative of the function. The antiderivative of the function is \frac{4}{3} x ^{\frac{3}{2}} - \frac{x^3}{3}

Use the fundamental theorem of calculus:

\int _{a} ^ {b} f(x) = [F(x)]_{a} ^{b} = F (b) - F(a)

Substitute \sqrt[3]{4} and 0 in the antiderivative of the function will give us the following value of area:

A = \frac{4}{3} u^2

Example 2

Find the area between two curves y = x^2 and y = 4x - x^2.

Solution

Follow these steps to calculate the area.

Step 1 - Sketch the graph

The graph of the two curves is given below:

Example 2 - Graph depicting the area between two curves

Step 2 - Find the boundaries

Calculate the boundaries of the function by equation both the equations like this:

x^2 = 4x - x^2

4x - 2x^2 = 0

2x(2 - x) = 0

x = 0 or x = 2

Hence, the boundaries of the function are 0 and 2.

Step 3 - Calculate the definite integral

To calculate the definite integral, first, use the information from the previous steps to write the functions in the following form:

A = \int^b_a [g(x) - f(x)] dx

A = \int^{2}_0 (4x - x^2 - x^2)dx

A = \int^2_0 (4x - 2x^2)dx

Find the antiderivative of the function. The antiderivative of the function is 2x^2 - \frac{2x^3}{3}

Use the fundamental theorem of calculus:

\int _{a} ^ {b} f(x) = [F(x)]_{a} ^{b} = F (b) - F(a)

Substitute 2 and 0 in the antiderivative of the function:

A = 2 (2)^2 - \frac{2 (2)^3}{3} - 0

A = \frac{8}{3} u^2

 

Need a Maths teacher?

Did you like the article?

1 Star2 Stars3 Stars4 Stars5 Stars 5.00/5 - 1 vote(s)
Loading...

Rafia Shabbir