The area between a curve and the y-axis can be found using two main methods. The first method consists of using rectangles and subtracting their areas so that we obtain the required area. The second method consists of expressing x as a function of y to obtain a definite integral.
In this article, we will learn about the methods we can use to find the area between a curve and the y-axis. Then, we will use these methods to solve some practice exercises.
Finding the area between a curve and the y-axis using rectangles
Consider the area shown in the following diagram. This area is enclosed by the curve $latex y=f(x)$ and the y-axis between $latex y_{1}=f(a)$ and $latex y_{2}=f(b)$.
We can find this area using rectangles. Therefore, using the diagram, we can determine that the area $latex A$ is equal to:
$latex A=$(Area of OTUV)$latex -$(Area of OPQR)$latex -$(Area under $latex f(x)$ between $latex x=a$ and $latex x=b$)
Thus, if we simplify this, we have the following formula:
$$A=bf(b)-af(a)-\int_{a}^{b} f(x)dx$$
Finding the area between a curve and the y-axis by expressing x as a function of y
The area between a curve and the y-axis can be calculated by expressing x as a function of y. This means that we have to evaluate the following:
$$A=\int_{f(a)}^{f(b)} xdy$$
Therefore, we consider the following
- $latex dy$ indicates that the limits $latex f(a) $ and $latex f(b)$ are limits of y.
- $latex f(a)$ is the lower limit and is found by using the value of $latex a$ in $latex f(x)$. That is, we find the value of $latex y_{1}$ from the diagram above.
- $latex f(b)$ is the upper limit and is found by using the value of $latex b$ in $latex f(x)$. That is, we find the value of $latex y_{2}$.
- $latex x $ is a function of $latex y$. We find this function by solving the original function for $latex x$. For example, $latex y=3x+1$ becomes $latex x=\frac{y-1}{3}$.
Then, we can evaluate the definite integral by following the steps seen in this article.
Area between a curve and the y-axis – Examples with answers
EXAMPLE 1
Find the area between the curve $latex y=x^2$ and the y-axis from $latex y=1$ to $latex y=4$. Use rectangles to solve.
Solution
To solve this using rectangles, we can observe that the area A is given by:
$$A=(4\times 2)-(1\times 1)-\int_{1}^{2} x^2 dx$$
The limits of the integral $latex x=1$ and $latex x=2$ are the equivalent of $latex y=1$ and $latex y=4$ respectively.
Finding the value of the definite integral, we have:
$$\int_{1}^{2} x^2 dx=\left[ \frac{x^3}{3} \right]_{1}^{2}$$
$$\int_{1}^{2} x^2 dx=\left[ \frac{8}{3} \right]-\left[ \frac{1}{3} \right]$$
$$\int_{1}^{2} x^2 dx= \frac{7}{3}$$
Therefore, we have:
$$A=8-1-\frac{7}{3}$$
$$A=\frac{14}{3}$$
EXAMPLE 2
Use the second method to find the area between the curve $latex y=x^2$ and the y-axis from $latex y=1$ to $latex y=4$.
Solution
To use the second method, we have to evaluate the following definite integral:
$$A=\int_{1}^{4} x dy$$
Then, we have to find x as a function of y. Therefore, we have:
$latex y=x^2$
$latex \sqrt{y}=x$
Now, we substitute and solve the integral:
$$\int_{1}^{4} \sqrt{y} dy=\int_{1}^{4} y^{\frac{1}{2}} dy$$
$$=\left[\frac{2y^{\frac{3}{2}}}{3}\right]_{1}^{4}$$
$$=\left[\frac{2(4)^{\frac{3}{2}}}{3}\right]-\left[\frac{2(1)^{\frac{3}{2}}}{3}\right]$$
$$=\frac{16}{3}-\frac{2}{3}$$
$$A=\frac{14}{3}$$
We can see that we obtained the same answer as in the previous example, so both methods are valid.
EXAMPLE 3
Find the area between the curve $latex y=x^2+1$ and the y-axis from $latex y=2$ to $latex y=5$ using the first method.
Solution
We can see that the required area is given by:
$$A=(5\times 2)-(2\times 1)-\int_{1}^{2} x^2+1 dx$$
Now, let’s find the value of the definite integral:
$$\int_{1}^{2} x^2+1 dx=\left[ \frac{x^3}{3}+x \right]_{1}^{2}$$
$$=\left[ \frac{8}{3} +2 \right]-\left[ \frac{1}{3} +1 \right]$$
$$= \frac{10}{3}$$
Therefore, we have:
$$A=10-2-\frac{10}{3}$$
$$A=\frac{14}{3}$$
EXAMPLE 4
Determine the area between the curve $latex y=x^2+1$ and the y-axis from $latex y=2$ to $latex y=5$ using the second method.
Solution
To solve the example using the second method, we have to solve the following integral:
$$A=\int_{2}^{5} x dy$$
Now, let’s find an equation of x as a function of y. Then, we have:
$latex y=x^2+1$
$latex \sqrt{y-1}=x$
Substituting this expression and solving the definite integral, we have:
$$\int_{2}^{5} \sqrt{y-1} dy=\int_{2}^{5} (y-1)^{\frac{1}{2}} dy$$
$$=\left[\frac{2}{3}(y-1)^{\frac{3}{2}}\right]_{2}^{5}$$
$$=\left[\frac{2}{3}(5-1)^{\frac{3}{2}}\right]-\left[\frac{2}{3}(2-1)^{\frac{3}{2}}\right]$$
$$=\left[\frac{2}{3}(4)^{\frac{3}{2}}\right]-\left[\frac{2}{3}(1)^{\frac{3}{2}}\right]$$
$$=\left[\frac{2}{3}(8)\right]-\left[\frac{2}{3}(1)\right]$$
$$=\frac{16}{3}-\frac{2}{3}$$
$$A=\frac{14}{3}$$
EXAMPLE 5
Find the area between the curve $latex y=x^3+1$ and the y-axis from $latex y=1$ to $latex y=4$ using any method.
Solución
In this case, method 2 may be simpler, so we are going to use that method. Then, we have to evaluate the following definite integral:
$$A=\int_{1}^{4} x dy$$
Now, we find an equation for x in terms of y:
$latex y=x^3+1$
$latex \sqrt[3]{y-1}=x$
We can substitute the expression found and solve the definite integral:
$$\int_{1}^{4} \sqrt[3]{y-1} dy=\int_{1}^{4} (y-1)^{\frac{1}{3}} dy$$
$$=\left[\frac{3}{4}(y-1)^{\frac{4}{3}}\right]_{1}^{4}$$
$$=\left[\frac{3}{4}(4-1)^{\frac{4}{3}}\right]-\left[\frac{3}{4}(1-1)^{\frac{4}{3}}\right]$$
$$=\left[\frac{3}{4}(4.327)\right]-[0]$$
$$A=3.245$$
Area between a curve and the y axis – Practice problems
See also
Interested in learning more about integrals of functions? You can take a look at these pages:
Jefferson Huera Guzman
Jefferson is the lead author and administrator of Neurochispas.com. The interactive Mathematics and Physics content that I have created has helped many students.
