Finding Areas by Integration

Expert reviewed • 22 November 2024 • 10 minute read

HSC Maths Advanced Syllabus

use the notation of the definite integral $\int_a^bf(x)dx$ for the area under the curve $y=f(x)$ from $x=a$ to $x=b$ if $f(x)\geq0$
use geometric ideas to find the definite integral $\int_a^bf(x)dx$ where $f(x)$ is positive throughout an interval $a\leq x \leq b$ and the shape of $f(x)$ allows such calculations, for example when $f(x)$ is a straight line in the interval or $f(x)$ is a semicircle in the interval
calculate the area under a curve
calculate areas between curves determined by any functions within the scope of this syllabus
Integrate functions and find indefinite or definite integrals and apply this technique to solving practical problems

Note:

Video coming soon!

How to Find Area by Integration

Integration is a highly accurate method of determining the area under a curve on a graph. To find the area we must first be given an integral that is or has the potential to be governed by limits. If we have this, we can follow the steps below to solve the area.

Draw the graph and display relevant date, ie. intercepts (see module 3 for specific instructions)
Create/determine the necessary definite integrals that correspond to a specific area.
Evaluate the integral, and display the answer as area with square units.

Finding the Area in Regards to the Y-Axis

So far all the examples we have explored, refer to finding the area between a curve and the x-axis. However, we can do the same for the y-axis. The process behind determining an area between a curve and the y-axis, is the same as usual, however, we use different notation. We say that for a function governed by the limits $a\leq y\leq b$ , the definite integral notation is:

\int_a^bf(y)dy\qquad or\qquad \int_a^bx.dy

Practice Question 1

Find the area created between the $y$ axis and the curve $y=x+1$ between the limits $2\leq y\leq6$

First we must sketch the curve $y=x+1$ and $y=6$ .

Now we must create an integral to determine the required area. However, we must first manipulate the equation to make $x$ the subject.

y=x+1\\x=y-1

Now we can create an integral and evaluate:

\int_2^6(y-1)dy=[\frac{1}{1+1}y^{1+1}-\frac{1}{0+1}y^{0+1}]_2^6\\=[\frac{y^2}{2}-y]_2^6\\=(\frac{6^2}{2}-6)-(\frac{2^2}{2}-2)\\=(18-6)-0\\=12 units^2

Area of Compound Regions and Area’s Between Curves

So far in this chapter, all we have looked at is the area between a curve and an axis. Now we learn how to calculate the area formed between two different curves. Determining the area between two different curves generally uses a very similar process to that of solving an area between one curve and an axis. However, there is a slightly different method in the process of calculating the area between two curves. Before integrating, we must determine the greater function. This is the curve which is graphically "higher". After doing this we can use the following formula to determine the area, where the function $g(x)>f(x)$ :

\int_a^b(g(x)-f(x))dx

We can use a similar concept when calculating the area between two compound regions. When doing this, it is easier to create two expressions and add multiple areas together. For example, in the following graph, the easiest method to calculate the area, is by creating two expressions and summing both areas.

Practice Question 2

Determine the area of the region formed between the curves $y=x$ and $y=x^2$ . Do so between limits $0\leq x\leq1$ . The graph of these curves is provided below:

From the graph we can see what area we are finding. We also know our limits and equations from the question. As such, we must determine which function is greater. By visually inspecting the graph, we can see that $y=x$ is greater.

Now we must create a corresponding integral and solve the area using the formula in the chapter above.

\int_0^1((x)-(x^2)dx=\int_0^1(x-x^2)dx\\=[\frac{1}{1+1}x^{1+1}-\frac{1}{2+1}x^{2+1}]_0^1\\=[\frac{x^2}{2}-\frac{x^3}{3}]_0^1\\=(\frac{1^2}{2}-\frac{1^3}{3})-(\frac{0^2}{2}-\frac{0^3}{3})\\=(\frac{1}{2}-\frac{1}{3})-0\\=\frac{1}{6}units^2

The Indefinite Integral

Up Next

The Trapezoidal Rule

Return to Module 4: Integration