Area Under a Curve

How to find the area under curves using definite integrals; tutorials, with examples and detailed solutions are presented . A set of exercises with answers is presented at the bottom of the page. Also tutorials on area between curves is included.

Area under a curve

Approximation of area under a curve

Figure 1. Approximation of area under a curve by the sum of areas of rectangles.

We may approximate the area under the curve from x = x 1 to x = x n by dividing the whole area into rectangles. For example the area first rectangle (in black) is given by: area of rectangle
and then add the areas of these rectangles as follows: approximate formula for area under curve
If Δx in the above approximation of the area expression becomes small enough, the sum of the areas of the rectangles will approaches the exact value of the area under the curve. Hence we define the area as follows: Area Formula as a Sum
The above limit exists, it has the following notation using the concept of definite integrals.
Area Formula as an Integral
We now present several examples on how to use integrals to find the area under a curve. Detailed solutions to these examples are also included.

Examples with Detailed Solutions

Example 1

Find the area of the region bounded by y = 2x, y = 0, x = 0 and x = 2.(see figure below).

area under curve, example 1, triangle

Figure 2. Area under a curve example 1.

Solution to Example 1

Two methods are used to find the area. \( \) \( \) \( \) \( \)
Method 1 This problem may be solved using the formula for the area of a triangle.
\( \text{area} = (1/2) \times \text{base} \times \text{height} = (1/2) \times 2 \times 4 = 4 \; unit^2 \)

Method 2
We shall now use definite integrals to find the area defined above. If we let f(x) = 2x , using the formula of the area given by the definite integral above, we
\( \displaystyle \text{Area} = \int_{0}^{2} (2x) dx = 2 \int_{0}^{2} x dx = [2(x^2/2)]_0^2 = 4 \; \text{unit}^2 \)
The first method is fast but works because the area is that of a triangle, however the second method work for figures other than triangles.




Example 2

Find the area of the region bounded by y = 0.1 x3, y = 0, x = 2 and x = 4.

Solution to Example 2


 We first graph the given function and identify the region whose area is to be found.
area under curve, example 2, 0.1 x^3

Figure 3. Area under a curve example 2 , y = 0.1 x3, x = 2 , x = 4 and y = 0.


 Use the definite integrals to find the area as follows:.
\( \displaystyle \text{Area} = \int_{2}^{4} (0.1 x^3) dx = 0.1 \int_{2}^{4} x^3 dx = 0.1 \left [x^4/4 \right ]_2^4 \\\\ = 0.1 [4^4/4 - 2^4/4] = 6 \;\text{unit}^2 \)




Example 3

Find the area of the finite region bounded by the curve of \( y = 3(x - 1)(x - 3) \) and the x axis.

Solution to Example 3

Note that the limits of integration are not given and therefore a detailed study of the graph of the given function is necessary. The graph of the given function shows that there are two x intercept that are easy to find since the given function is in factored form: x = 1 and x = 2. The finite region is bounded by the curve of y = 3(x - 1)(x - 3), x = 1, x = 3 and the x axis as shown below in the graph.
finite area between curve and x axis, example 3

Figure 4. Finite area between the curve example 3, the x intercepts and the x axis (y = 0)

 Use the definite integrals to find the area as follows:
\( \displaystyle \int_{1}^{3} (3(x - 1)(x - 3)) dx = 3 \int_{1}^{3} (x^2 - 4x + 3) dx = 3 \left [x^3 / 3 - 4 x^2/2 + 3x \right ]_1^3 \\\\ = 3 \left [ (3)^3 / 3 - 4 (3)^2/2 + 3(3) - ( (1)^3 / 3 - 4 (1)^2/2 + 3(1) ) \right ] = - 4 \)

Note that the definite integral found is negative and that is because y = 3(x - 1)(x - 3) is negative between the limits of integration x = 1 and x = 3. The area is the absolute value of -4 and is therefore 4 unit2.




Example 4

Find the area of the finite region bounded by the curve of \( y = - 0.25 x (x + 2)(x - 1)(x - 4) \) and the x axis.

Solution to Example 4

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The given function is a polynomial of degree 4 with negative leading coefficient. We graph the given function and study it in order to identify the finite region bounded by the curve and x axis. The graph of the given function has 3 x-intercepts: x = - 2, 1 and 4. The finite region is composed of three regions. The first one from x = - 2 to x = 0. The second region from x = 0 to x = 1 and the third from x = 1 to x = 4.
finite area between curve and x axis, example 4

Figure 5. Finite area between the curve and the x axis in example 4
Let us calculate the following definite integrals taking as limits the x intercepts.
Region 1
\(\displaystyle I_1 = \int_{-2}^{0} (- 0.25 x (x + 2)(x - 1)(x - 4)) dx \)

Expand \( x (x + 2)(x - 1)(x - 4) \) and move - 0.25 outside the operation of integration.
\(\displaystyle = -0.25 \int_{-2}^{0} (x^4-3x^3-6x^2+8x) dx \\\\ = - 0.25 \left [ x^5/5-3x^4/4-6x^3/3+8x^2/2 \right ]_{-2}^0 = 3.4 \)

Region 2
\(\displaystyle I_2 = \int_{0}^{1} (- 0.25 x (x + 2)(x - 1)(x - 4)) dx \\\\ = -0.25 \int_{0}^{1} (x^4-3x^3-6x^2+8x) dx \\\\ = - 0.25 \left [ x^5/5-3x^4/4-6x^3/3+8x^2/2 \right ]_{0}^1 = -0.3625 \)

Region 3
\(\displaystyle I_3 =\int_{1}^{4} (- 0.25 x (x + 2)(x - 1)(x - 4)) dx \\\\ = -0.25 \int_{1}^{4} (x^4-3x^3-6x^2+8x) dx \\\\ = - 0.25 \left [ x^5/5-3x^4/4-6x^3/3+8x^2/2 \right ]_{1}^4 = 13.1625 \)

Note that I2 is negative because between x = 0 and x = 1 , \( y = - 0.25 x (x + 2)(x - 1)(x - 4) \) is negative. Hence we need to take the absolute value of the definite integral of I2 in order to find the area of the region from x = 0 to x = 1. Hence the total area is given by:

\( Area = I_1 + | I_2 | + I_3 = 3.4 + |-0.3625| + 13.1625 = 16.925 \; \text{unit}^2 \)




Example 5

Find k so that the area of the finite region bounded by the curve of \( y = - x( x - k) \) and the x axis is equal to 4/3 units2.

Solution to Example 5

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The graph of the given function is a parabola that opens downward and has two x intercepts: x = 0 and x = k. The finite region bounded by the curve and the x axis is limited at the x intercepts as shown in the graph below.
Find unknown k given area between curve and x axis, example 5

Figure 6. Finite area between parabola and x axis in example 5
The area between the curve and y = 0 is given by
\(\displaystyle \text{Area} = \int_{0}^{k} (- x( x - k)) dx \\\\ \)

Expand - x( x - k)
\(\displaystyle = \int_{-2}^{0} (- x^2 + kx ) dx = \left [ - x^3/3 + k x^2/2 \right ]_{0}^k = -k^3 / 3 + k^3 / 2 = k^3 / 6 \)

As expected, the expression for the area includes the parameter k which is calculated by setting the area equal to 4/3. Hence
\( k^3 / 6 = 4 / 3 \)
Solve the above equation for k to obtain
\( k = 2 \)



Exercises


1) Find the area of the finite region enclosed by \( y = -(x+1)(x-3) \) and y = 0
2) Find the area of the finite region bounded b \( y = sin(x) \), y = 0 , x = 0 and x = 2π.
3) Find k positive so that the area under the curve of \( y = (x + 2) \) the vertical axes x = 0, x = k and the x axis is equal to 2.

Answers to Above Exercises


1) 32/3
2) 4
3) k = 2e 2 - 2

More References and links

integrals and their applications in calculus.
Area between two curves .
Volume of a Solid of Revolution .