# 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 curveWe 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:
y \codt \Delta x = f(x_1) \Delta x
and then add the areas of these rectangles as follows:
\text{Approximate Area} = \sum_{i=1}^{n-1} f(x_i) \Delta x
\text{Exact Area} = \lim_{\Delta x\to\0} \sum_{i=1}^{n-1} f(x_i) \Delta x
\text{Exact Area} = \lim_{\Delta x\to\0} \sum_{i=1}^{n-1} f(x_i) \Delta x = \color{red}{\int_{x_1}^{x_2} f(x) dx}
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. ## Example 1Find the area of the region bounded by y = 2x, y = 0, x = 0 and x = 2.(see figure below).## Solution to Example 1Two methods are used to find the area.Method 1 This problem may be solved using the formula for the area of a triangle.
area = (1/2) × base × height = (1/2)× 2 × 4 = 4 unit ^{2}Method 2We 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
\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 2Find the area of the region bounded by y = 0.1 x^{3}, y = 0, x = 2 and x = 4.
## Solution to Example 2We first graph the given function and identify the region whose area is to be found. Use the definite integrals to find the area as follows:.
\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 3Find the area of the finite region bounded by the curve of y = 3(x - 1)(x - 3) and the x axis.## Solution to Example 3Note 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.Use the definite integrals to find the area as follows:
\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 ]_2^4 = - 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 unit ^{2}.## Example 4Find 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.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.Let us calculate the following definite integrals taking as limits the x intercepts.
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.
= -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
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
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 I _{2} 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 I_{2} 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 5Find 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 units^{2}.
## Solution to Example 5.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.The area between the curve and y = 0 is given by
\text{Area} = \int_{0}^{k} (- x( x - k)) dx \\\\
Expand - x( x - k)
= \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
## Exercises1) 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 below. 3) Find k positive so that the area under the curve of y = (x + 2), x = 0, x = k and the x axis is equal to 2 ## Answers to Above Exercises1) 32/3 2) 4 3) k = 2e ^{2} - 2
## More References and linksintegrals and their applications in calculus.Area between two curves. Volume of a Solid of Revolution. |