Area Under Curve

Area Under Curve is area enclosed by curve and the coordinate axes, it is calculated by taking very small rectangles and then taking their sum if we take infinitely small rectangles then their sum is calculated by taking the limit of the function so formed.

For a given function f(x) defined in the interval [a, b], the area (A) under the curve of f(x) from 'a' to 'b' is given by A = ∫_a ^b f(x)dx. The area under a curve is computed by taking the absolute value of the function over the interval [a, b], summed over the range.

In this article, we will learn about, the area under the curve, its applications, examples, and others in detail.

What is Area Under Curve?

Area Under the Curve is area enclosed by any curve with the x-axis and given boundary conditions i.e., the area bounded by function y = f(x), x-axis, and the line x = a, and x = b. In some cases, there is only one or no boundary condition as the curve intersects the x-axis either once or twice respectively.

Area under the curve can be calculated using various methods such as Reimann sum, and Definite integral and we can also approximate the area using the basic shapes i.e., triangle, rectangle, trapezium, etc.

Read in Detail:

• Calculus in Maths
• Integration

Calculating the Area Under the Curve

To calculate area under a curve, we can use the following methods such as:

• Using Reimann Sums
• Using Definite Integrals
• Using Approximation

Let's study these methods in detail as follows:

Using Reimann Sums

Reimann Sums is calculated by dividing a given function's graph into smaller rectangles and summing the areas of each rectangle. The more rectangles we consider by subdividing the provided interval, the more precise the area computed by this approach is; nevertheless, the more subintervals we consider, the more difficult the calculations get.

Reimann Sum can be classified into three more categories such as:

• Left Reimann Sum
• Right Reimann Sum
• Midpoint Reimann Sum

Area using the Reimann sum is given as follows:

\bold{Area = \sum_{i=1}^{n}f(x_i)\Delta x_i}

where,

• f(x_i) is the value of the function being integrated at the i^th sample point
• Δx = (b-a)/n is the width of each subinterval, 
  • a and b are the limits of integration and 
  • n is the number of subintervals
• ∑ represents the sum of all the terms from i=1 to n, 

Example: Find the area under the curve for function, f(x) = x² between the limits x = 0 and x = 2.

Solution:

We want to find the area under the curve of this function between x = 0 and x = 2. We will use a left Reimann Sum with n = 4 subintervals to approximate the area.

Let's calculate the area under the curve using 4 subintervals. 

Thus, width of subintervals, Δx = (2-0)/4 = 0.5

All the 4 subintervals are,

a = 0 = x₀ < x₁ <x₂ < x₃ < x₄ = 2 = b

x₀ = 0, x₁ = 0.5, x₂ = 1, x₃ = 1.5, x₄ = 2

Now we can evaluate the function at these x-values to find the heights of each rectangle:

f(x₀) = (0)² = 0
f(x₁) = (0.5)² = 0.25
f(x₂) = (1)² = 1
f(x₃) = (1.5)² = 2.25
f(x₄) = (2)² = 4

Area under the curve can now be approximated by summing the areas of the rectangles formed by these heights:

A ≈ Δx[f(x₀) + f(x₁) + f(x₂) + f(x₃)] = 0.5[0 + 0.25 + 1 + 2.25] = 1.25

Therefore, area under the curve of f(x) = x² between x = 0 and x = 2, approximated using a left Reimann Sum with 4 subintervals, is approximately 1.25.

Using Definite Integrals

Definite Integral is the almost same as the Reimann sum but here the number of subintervals approaches infinity. If the function is given for interval [a, b] then definite integral is defined as:

 \int_{a}^{b} f(x) dx = \lim_{n\to \infty}\sum_{i=1}^{n}f(x_i)\Delta x_i

A definite Integral gives the exact area under the curve, unlike the Reimann sum.  The definite integral is calculated by finding the antiderivative of the function and evaluating it at the limits of integration.

Area with Respect to X-Axis

Curve shown in the image below is represented using y = f(x). We need to calculate the area under the curve with respect to the x-axis. The boundary values for the curve on the x-axis are a and b respectively. The area A under this curve with respect to the x-axis is calculated between the points x = a  and x = b. Consider the following curve:

Formula for Area Under the Curve w.r.t to the x-axis is given by:

\bold{A = \int_{a}^{b}y.dx}

\bold{A = \int_{a}^{b}f(x)dx}

where,

• A is Area Under Curve
• y or f(x) is Equation of Curve
• a, and b are x-values or limit of integration, for which we need to calculate area

Area with Respect to Y-Axis

Curve shown in the image above is represented using x = f(y). We need to calculate the area under the curve with respect to Y-axis. The boundary values for the curve on the Y-axis are a and b respectively. The area A under this curve with respect to Y-axis between the points y = a and y = b. Consider the following curve:

Formula for Area Under the Curve w.r.t to the y-axis is given by:

\bold{A = \int_{a}^{b}x.dy}

\bold{A = \int_{a}^{b}f(y)dy}

where,

• A is Area Under Curve
• x or f(y) is Equation of Curve
• a, b are y-Intercepts

Articles related to Area Under Curve:

• Area Between Two Curves
• Area as Definite Integral
• Area between Polar Curves
• Area under Simple Curves

Approximating Area Under Curve

Approximating the area under the curve involves using simple geometric shapes, such as rectangles or trapezoids, to estimate the area under the curve. This method is useful when the function is difficult to integrate or when it is not possible to find an antiderivative of the function. The accuracy of the approximation depends on the size and number of the shapes used.

Calculating Area Under Curve

We can easily calculate the area of the various curve using the concepts discussed in the given article. Now let's consider some examples of calculation of Area Under the Curve for some common curves.

Area Under Curve: Parabola

We know that a standard parabola is divided into two symmetric parts by either the x-axis or the y-axis. Suppose we take a parabola y² = 4ax and then its area is to be calculated from x = 0 to x = a. And if required we double its area to find the area of the parabola in both the quadrant.

Calculating area,

y² = 4ax

y = √(4ax)

A = 2∫₀^a y.dx

A = 2∫₀^a√(4ax).dx

A = 4√(a)∫₀^a√(x).dx

A = 4√(a){2/3.a^3/2}

A = 8/3a²

Thus, area under the parabola from x = 0 to x = a is 8/3a² square units

Area Under Curve: Circle

A circle is a closed curve whose circumference is always at an equal distance from its center. Its area is calculated by first calculating the area in the first quadrant and then multiplying it by 4 for all four quadrants.

Suppose we take a circle x² + y² = a² and then its area is to be calculated from x = 0 to x = a in the first quadrant. And if required we quadruple its area to find the area of the circle.

Calculating area,

x² + y² = a²

y = √(a² - x²).dx

A = 4∫₀^a y.dx

A = 4∫₀^a√(a² - x²).dx

A = 4[x/2√(a² - x²) + a²/2 sin^-1(x/a)]^a₀

A = 4[{(a/2).0 + a²/2.sin^-1} - 0]

A = 4(a²/2)(π/2)

A = πa²

Thus, area under the circle is πa² square units

Area Under Curve: Ellipse

A circle is a closed curve. Its area is calculated by first calculating the area in the first quadrant and then multiplying it by 4 for all four quadrants.

Suppose we take a circle (x/a)² + (y/b)² = 1 and then its area is to be calculated from x = 0 to x = a in the first quadrant. And if required we quadruple its area to find the area of the ellipse.

Calculating area,

(x/a)² + (y/b)² = 1

y = b/a√(a² - x²).dx

A = 4∫₀^a y.dx

A = 4b/a∫₀^a√(a² - x²).dx

A = 4b/a[x/2√(a² - x²) + a²/2 sin^-1(x/a)]^a₀

A = 4b/a[{(a/2).0 + a²/2.sin^-1} - 0]

A = 4b/a(a²/2)(π/2)

A = πab

Thus, the area under the ellipse is πab square units.

Area Under Curve Formulae

Formula for various types of calculation of Area Under Curve is tabulated below:

Sample Examples on Area Under Curve

Example 1: Find the area under the curve y² = 12x and the X-axis.

Solution:

Given curve equation is y² = 12x

This is an equation of parabola with a = 3 so, y² = 4(3)(x)

Graph for the required area is shown below:

X-axis divides the above parabola into 2 equal parts. So, we can find the area in the first quadrant and then multiply it by 2 to get the required area

So, we can find the required area as:

A = 2\int_{a}^{b}ydx

⇒ A = 2\int_{0}^{3}\sqrt{12x}dx

⇒ A = 2\sqrt{12}[\frac{2x^\frac{3}{2}}{3}]_0^3

⇒ A = \frac{4\sqrt{12}}{3}[x^\frac{3}{2}]_0^3

⇒ A = \frac{4\sqrt{12}}{3}*\sqrt{27}

⇒ A = 24 sq. units

Example 2: Calculate the area under the curve x = y³ - 9 between the points y = 3 and y = 4.

Solution:

Given, equation of curve is x = y³ - 9

Boundary points are (0, 3) and (0, 4)

As the equation of curve is of the form x = f(y) and the points are also on the Y-axis, we will use the formula,

A = \int_{a}^{b}x.dy

⇒ A = \int_{3}^{4}(y^3-9)dy

⇒ A = [\frac{y^4}{4}-9y]^4_3

⇒ A = (64-36)-(\frac{81}{4}-27)

⇒ A = 28+\frac{27}{4}

⇒ A = 139/4 sq. units

Example 3: Calculate the area under the curve y = x² - 7 between the points x = 5 and x = 10.

Solution:

Given, the curve is y = x²−7 and the boundary points are (5, 0) and (10, 0)

Thus, area under the curve is given by: 

A = \int_{5}^{10}(x^2-7)dx

⇒ A = [\frac{x^3}{3}-7x]_5^{10}

⇒ A = (100/3 - 70) - (125/3 - 35)

⇒ A = 790/3 - 23/3

⇒ A = 770/3 sq. units

Example 4: Find the area enclosed by the parabola y² = 4ax and the line x = a in the first quadrant.

Solution:

Curve and the line given can be drawn as follows:

Now, the equation of curve is y² = 4ax

Boundary points come out to be (0, 0) and (a, 0)

So the area with respect to X-axis can be calculated as:

A=\int_{0}^{a}ydx

⇒ A=\int_{0}^{a}\sqrt{4ax}dx

⇒ A=[\sqrt{4a}\frac{x^{\frac{1}{2}+1}}{\frac{3}{2}}]_0^a

⇒ A=2×\frac{2}{3}\sqrt{a}[x^{\frac{3}{2}}]_0^a

⇒ A=\frac{4\sqrt{a}}{3}×a^\frac{3}{2}

⇒ A=\frac{4a^2}{3} sq. units

Example 5: Find the area covered by the circle x² + y² = 25 in the first quadrant.

Solution:

Given, x² + y² = 25

Curve can be drawn as:

Required area has been shaded in the above figure. From the equation we can see that radius of the circle is 5 units.

As, x² + y² = 25 

y = \sqrt{25-x^2}

To find the area, we shall use:

A = \int_{a}^{b}ydx

⇒ A = \int_{0}^{5}\sqrt{25-x^2}dx

⇒ A = [\frac{x}{2}(\sqrt{25-x^2}+\frac{25}{2}sin^{-1}\frac{x}{5})]_0^5

⇒ A = [(\frac{5}{2}×0 +\frac{25}{2}sin^{-1}(1))-0]

⇒ A = \frac{25}{2}×\frac{\pi}{2}

⇒ A = 25 π/4 sq. units