In this article, we will discuss how to calculate the area between a function and the x-axis.

Area between a Function and the x-axis

There are two scenarios in which we can find the area between a function and the x-axis:

  • When the function is non-negative, i.e. positive
  • When the function is negative

Now, let us see what steps one should follow while calculating the area between the function and x-axis.

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Steps for Calculating Area Under the Curve

To compute the area under the curve f(x), one should follow the steps below:

Step 1 - Sketch the area

Step 2 - Find the boundaries a and b

Step 3 - Write the definite integral function

Step 4 - Integrate the function

In the next section, we will discuss how to calculate the area when the function is positive.

When the Function is Positive

If the function f(x) is positive on an interval [a, b], then the area of the region bounded by the graph of the function f, the horizontal axis (x - axis) and the vertical lines x = a, and x = b is defined as follows:

A = \int_a ^{b} f(x) dx

 

Now, let us solve some examples in which we will calculate the area under the curve.

Example 1

Calculate the area bounded by the function f(x) = 9 - x^2 and the x-axis.

Solution

Follow these steps to solve this example.

Step 1 - Sketch the graph

Sketch the graph. The graph of the function f(x) = 9 - x^2 is given below:

Example 1 - Graph of the function y = 9 - x^2

Step 2 - Find the boundaries

To find the boundaries a and b of the function, we have to determine the x-intercepts of the curve. Although, from the graph above, we can clearly determine that the limits of integration are 3 and -3, however, we can check it again by equating the function f(x) = 9 - x^2 equal to zero as shown below:

9 - x^2 = 0

9 = x^2

x = \pm 3

Step 3 - Using the boundaries, write the definite integral function

In this step, we will use the boundaries 3 and -3 and write the function in definite integral form like this:

A = \int _{-3}^{3} (9 - x^2) dx

Step 4 - Integrate the function

Write the function A = \int _{-3}^{3} (9 - x^2) dx using the sum/difference property of definite integrals like this:

A = \int^{3}_{-3} 9 f(x) dx - \int^{3}_{-3} x^2 g(x) dx

Now, to find the definite integral of the function, first, we will calculate the antiderivative of the function. The antiderivative of the function is 9x - \frac{x^3}{3} + C. For ease of calculation, consider C = 0.

The fundamental theorem of calculus says that:

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

Substitute 3 and -3 in the antiderivative of the function:

A = 9 (3) - \frac{3^3}{3} - 9(-3) - \frac{3^3}{3}

A = 27 - \frac{27}{3} - (-27 + \frac{27}{3})

= 18 - (-18)

= 18 + 18 = 36

Hence, the area bounded by the function f(x) = 9 - x^2 and the x- axis is 36 u^2

 

Example 2

Calculate the area of the region enclosed by the function xy = 36, the lines x = 6 and x = 12 and the x-axis.

Solution

Follow these steps to compute the area.

Step 1 - Sketch the graph

The graph of the function y = \frac{36}{x} is given below:

 

Example 2 - Graph of the function xy=3x

Step 2 - Find the boundaries

The boundaries x = 6 and x = 12 are already given in this problem.

Step 3 - Using the boundaries, write the definite integral function

In this step, we will use the boundaries 6 and 12 and write the function in definite integral form like this:

A = \int _{6}^{12} (\frac{36}{x}) dx

Step 4 - Integrate the function

Now, to find the definite integral of the function, first, we will compute the antiderivative of the function. The antiderivative of the function is 36lnx + C. For ease of calculation, consider C = 0.

The fundamental theorem of calculus says that:

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

Substitute 6 and 12 in the antiderivative of the function like this:

A = 36 ln (12) - 36 ln (6) = 36 ln 2 u^2

When the Function is Negative

If the function is negative in a closed interval [a, b] then the graph of the function is below the horizontal axis. The area of the function can be defined like this:

A = - \int ^b _ a f(x) dx

A = |\int ^b_a f(x) dx|

 

Now, let us solve some examples in which we are given a negative function.

Example 1

Find the area of the function bounded by the graph of  y = x^2 - 4x and x-axis.

Solution

Follow these steps to solve this example.

Step 1 - Sketch the graph

Sketch the graph. The graph of the function f(x) =  x^2 - 4x is given below:

Example 1 - Graph of the function x^2 - 4x

The downward parabola shows that the function is negative.

Step 2 - Find the boundaries

To find the boundaries a and b of the function, we have to determine the x-intercepts of the curve. Although, from the graph above, we can clearly determine that the limits of integration are 4 and 0, however, we can check it again by equating the function f(x) = x^2 - 4x equal to zero as shown below:

x^2 - 4x= 0

x = 4 or x = 0

Step 3 - Using the boundaries, write the definite integral function

In this step, we will use the boundaries 4 and 0 and write the function in definite integral form like this:

A = \int _{0}^{4} (x^2 - 4x) dx

Step 4 - Integrate the function

Write the function A = \int _{0}^{4} (x^2 - 4x) dx using the sum/difference property of definite integrals like this:

A = \int^{4}_{0} x^2 f(x) dx - \int^{4}_{0} (-4x) g(x) dx

Now, to find the definite integral of the function, first, we will calculate the antiderivative of the function. The antiderivative of the function is \frac{x^3}{3} - 2x^2+ C. For ease of calculation, consider C = 0.

The fundamental theorem of calculus says that:

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

Substitute 4 and 0 in the antiderivative of the function:

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

A = - \frac{32}{3} u^2

|A| = \frac{32}{3} u^2

Example 2

Calculate the area bounded by the curve y = cos x and the x-axis between \frac{\pi}{2} and \frac{3\pi}{2}.

Solution

Follow these steps to solve this example.

Step 1 - Sketch the graph

Sketch the graph. The graph of the function y = cos x is given below:

Example 2 - Graph of the function y = cos x

Step 2 - Find the boundaries

The boundaries x = \frac{3 \pi}{2} and x = \frac{\pi}{2} are already given in this problem.

Step 3 - Using the boundaries, write the definite integral function

In this step, we will use the boundaries x = \frac{3 \pi}{2} and x = \frac{\pi}{2} to write the function in definite integral form like this:

A = \int _{\frac{\pi}{2}}^{\frac{3\pi}{2}} cos x dx

Step 4 - Integrate the function

Now, to find the definite integral of the function, first, we will compute the antiderivative of the function. The antiderivative of the function is sin x + C. For ease of calculation, consider C = 0.

The fundamental theorem of calculus says that:

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

Substitute x = \frac{3 \pi}{2} and x = \frac{\pi}{2} in the antiderivative of the function like this:

A = sin \frac{3\pi}{2} - sin \frac{\pi}{2} = -1 -1 = -2

|A| = 2u^2

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Rafia Shabbir