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 provided at the bottom of the page. Tutorials on area between curves are also included.
Area Under a Curve: Review
We may approximate the area under the curve from \( x = x_1 \) to \( x = x_n \) by dividing the total area into rectangles.
For example, the area of the first rectangle is given by:
\[ y \Delta x = f(x_1)\Delta x \]Adding the areas of these rectangles gives:
\[ \text{Approximate Area} = \sum_{i=1}^{n-1} f(x_i)\Delta x \]If \( \Delta x \) in the above approximation becomes small enough, the sum of the areas of the rectangles will approach the exact value of the area under the curve.
Hence, we define the exact area using a limit:
\[ \text{Exact Area} = \lim_{\Delta x \to 0} \sum_{i=1}^{n-1} f(x_i)\Delta x \]Using the concept of definite integrals, this limit is denoted as:
We now present several examples on how to use integrals to find the area under a curve, complete with detailed solutions.
Examples with Detailed Solutions
Example 1
Find the area of the region bounded by \( y = 2x \), \( y = 0 \), \( x = 0 \), and \( x = 2 \).
Solution to Example 1
Method 1: Geometric formula for the area of a triangle
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 4 = 4 \; \text{unit}^2 \]Method 2: Using calculus (definite integrals)
Let \( f(x) = 2x \):
\[ \text{Area} = \int_{0}^{2} (2x) \, dx = 2 \int_{0}^{2} x \, dx = 2 \left[ \frac{x^2}{2} \right]_0^2 = 4 \; \text{unit}^2 \]The first method is fast for triangles, but calculus works for arbitrary curves.
Example 2
Find the area of the region bounded by \( y = 0.1x^3 \), \( y = 0 \), \( x = 2 \), and \( x = 4 \).
Solution to Example 2
Use definite integrals to find the area:
\[ \begin{aligned} \text{Area} &= \int_{2}^{4} (0.1x^3)\,dx \\ &= 0.1 \int_{2}^{4} x^3\,dx \\ &= 0.1 \left[ \frac{x^4}{4} \right]_2^4 \\ &= 0.1 \left( \frac{4^4}{4} - \frac{2^4}{4} \right) = 6 \; \text{unit}^2 \end{aligned} \]Example 3
Find the area of the finite region bounded by the curve \( y = 3(x - 1)(x - 3) \) and the x-axis.
Solution to Example 3
The limits of integration are the x-intercepts, which are clearly \( x = 1 \) and \( x = 3 \) from the factored form.
\[ \begin{aligned} \int_{1}^{3} 3(x - 1)(x - 3)\,dx &= 3 \int_{1}^{3} (x^2 - 4x + 3)\,dx \\ &= 3 \left[ \frac{x^3}{3} - 2x^2 + 3x \right]_1^3 \\ &= 3 \left[ (9 - 18 + 9) - (\frac{1}{3} - 2 + 3) \right] = -4 \end{aligned} \]Note that the definite integral is negative because the curve is below the x-axis between \( x = 1 \) and \( x = 3 \). The actual area is the absolute value: \( |-4| = 4 \; \text{unit}^2 \).
Example 4
Find the area of the finite region bounded by the curve \( y = -0.25x(x + 2)(x - 1)(x - 4) \) and the x-axis.
Solution to Example 4
The x-intercepts are \( x = -2 \), \( x = 1 \), and \( x = 4 \). The region is split into three parts: from \( -2 \) to \( 0 \), from \( 0 \) to \( 1 \), and from \( 1 \) to \( 4 \).
Region 1 (\( [-2, 0] \)):
\[ I_1 = -0.25 \int_{-2}^{0} (x^4 - 3x^3 - 6x^2 + 8x) \, dx = 3.4 \]Region 2 (\( [0, 1] \)):
\[ I_2 = -0.25 \int_{0}^{1} (x^4 - 3x^3 - 6x^2 + 8x) \, dx = -0.3625 \]Region 3 (\( [1, 4] \)):
\[ I_3 = -0.25 \int_{1}^{4} (x^4 - 3x^3 - 6x^2 + 8x) \, dx = 13.1625 \]Taking the absolute value of negative regions, total area is:
\[ \text{Area} = I_1 + |I_2| + I_3 = 3.4 + |-0.3625| + 13.1625 = 16.925 \; \text{unit}^2 \]Example 5
Find \( k \) so that the area of the finite region bounded by \( y = -x(x - k) \) and the x-axis is equal to \( 4/3 \) units\(^2\).
Solution to Example 5
The intercepts are at \( x = 0 \) and \( x = k \):
\[ \text{Area} = \int_{0}^{k} (-x^2 + kx) \, dx = \left[ -\frac{x^3}{3} + \frac{kx^2}{2} \right]_{0}^{k} = \frac{k^3}{6} \]Setting this equal to \( 4/3 \):
\[ \frac{k^3}{6} = \frac{4}{3} \implies k^3 = 8 \implies k = 2 \]Exercises
- Find the area of the finite region enclosed by \( y = -(x + 1)(x - 3) \) and \( y = 0 \).
- Find the area of the finite region bounded by \( y = \sin(x) \), \( y = 0 \), \( x = 0 \), and \( x = 2\pi \).
- Find \( k > 0 \) so that the area under the curve \( y = x + 2 \) between vertical axes \( x = 0 \), \( x = k \), and the x-axis is equal to \( 2 \).
Answers to Above Exercises
- \( \dfrac{32}{3} \)
- \( 4 \)
- \( k = 2\sqrt{2} - 2 \)