Integrals can be used to find the area between a curve and the \(x\)-axis. We can use this to calculate the total accumulation of a quantity in physics, economics and engineering. Use this resource to learn how to calculate area under the curve.
Finding the area under a curve
Consider a curve defined by \(y=f(x)\). We can find the area under the curve between \(x=a\) and \(x=b\) using a definite integral:
\[ \int_{a}^{b}f(x)dx \]
This is the area bordered by the lines \(x=a\), \(x=b\), \(y=0\) (the \(x\)-axis) and the graph of \(f(x)\).
The area under a curve might be above the axis (and therefore, positive) or below the axis (and therefore, negative).
If \(f(x)\geq0\) between \(x=a\) and \(x=b\), you would get a positive value for the definite integral.
If, \(f(x)\leq0\) between \(x=a\) and \(x=b\), you would get a negative value for the definite integral.
Regardless of whether the area is above or below the curve, the area under a curve is always given as a positive number.
Where \(f(x)\) crosses the \(x\)-axis between \(a\) and \(b\), the value of the definite integral may be positive or negative. However, the area under the curve is always taken as the sum of the modulus of the signed areas.
For example, the function \(f(x)=x^{3}-x^{2}-2x\) is plotted. The graph cuts the \(x\)-axis at \(-1\), \(0\) and \(2\). The integral for \(-1\geq x\geq0\) is positive and the integral for \(0\geq x\geq2\) is negative. The overall integral from \(-1\geq x\geq2\) is \(-2.25\), but the area under the curve is positive. See Example 1 for how to work this out.
Example 1 – finding the area under the curve
Given the function \(f(x)=x^{3}-x^{2}-2x\), find the area under the curve between \(x=-1\) and \(x=2\).
To calculate the area, we need to evaluate two integrals. One from \(-1\) to \(0\) the other from \(0\) to \(2\). The actual area would then be the sum of the absolute value of these integrals. That is:
\[\begin{align*} \textrm{Area} & = \left|\int_{-1}^{0}\left(x^{3}-x^{2}-2x\right)dx\right|+\left|\int_{0}^{2}\left(x^{3}-x^{2}-2x\right)dx\right|\\
& = \left|\left[\frac{1}{4}x^{4}-\frac{1}{3}x^{3}-x^{2}\right]_{x=-1}^{x=0}\right|+\left|\left[\frac{1}{4}x^{4}-\frac{1}{3}x^{3}-x^{2}\right]_{x=0}^{x=2}\right|\\
& = \left|0-\left(\frac{1}{4}\left(-1\right)^{4}-\frac{1}{3}\left(-1\right)^{3}-\left(-1\right)^{2}\right)\right|+\left|\frac{1}{4}\left(2\right)^{4}-\frac{1}{3}\left(2\right)^{3}-2^{2}\right|\\
& = \left|-\left(\frac{1}{4}+\frac{1}{3}-1\right)\right|+\left|4-\frac{8}{3}-4\right|\\
& = \left|\frac{1}{4}+\frac{1}{3}-1\right|+\frac{8}{3}\\
& = \frac{5}{12}+\frac{8}{3}\\
& = \frac{37}{12}\textrm{ square units}
\end{align*}\]
Given the function \(f(x)=4x-x^{2}\), find the area between the graph of the function and the \(x\)-axis.
First, sketch the graph of the function. To do this, we determine where the graph of the function cuts the \(x\)-axis. This gives us the limits of integration for the definite integral.
To solve for \(x\), we can factorise the function and let \(y=0\):
\[\begin{align*} 0 & = 4x-x^{2}\\
& = x(4-x)\\
0=x\quad & \textrm{ or }\quad0=4-x\\
x=0\quad & \textrm{ or }\quad x=4
\end{align*}\]
This means that the graph cuts the \(x\)-axis at \(x=0\) and \(x=4\). \(x=0\) becomes the lower limit for the integral and \(x=4\) becomes the upper limit. The graph is:
The function is above the \(x\)-axis, so the area will be positive. We do not need the absolute value sign.
\[\begin{align*} \textrm{Area} & = \int_{0}^{4}f(x)dx\\
& =\int_{0}^{4}(4x-x^{2})dx\\
& = \left[\frac{4x^{2}}{2}-\frac{x^{3}}{3}\right]_{x=0}^{x=4}\\
& = \frac{4(4)^{2}}{2}-\frac{(4)^{3}}{3}-\left(\frac{4(0)^{2}}{2}-\frac{(0)^{3}}{3}\right)\\
& =\frac{64}{2}-\frac{64}{3}\\
& =\frac{3(64)-2(64)}{6}\\
& =\frac{64}{6}\\
& =\frac{32}{3}\textrm{ square units}
\end{align*}\]
Given the function \(f(x)=x^{2}-1\), find the area bounded by the curve, the \(x\)-axis and the lines \(x=-1\) and \(x=1\).
We need to sketch the graph of the function and then consider the range of integration. Factorising the function and solving for \(x\) when \(y=0\):
\[\begin{align*} 0 & = x^{2}-1\\
& = (x+1)(x-1)\\
x+1=0\quad & \textrm{ or }x-1=0\\
x=-1\quad & \textrm{ or }x=1
\end{align*}\]
The \(x\)-axis intercepts where \(f(x)=x^{2}-1=0\) are \(x=-1\) and \(x=1\).
The area will be negative as \(f(x)\) is below the axis between the points of integration. So, we must use the modulus of the definite integral:
\[\begin{align*} \textrm{Area } & =\left|\int_{-1}^{1}(x^{2}-1)dx\right|\\
& =\left|\left[\frac{1}{3}x^{3}-x\right]_{x=-1}^{x=1}\right|\\
& =\left|\frac{1}{3}-1-\left(\frac{1}{3}(-1)^{3}-(-1)\right)\right|\\
& =\left|\frac{1}{3}-1-\left(-\frac{1}{3}+1\right)\right|\\
& =\left|-\frac{2}{3}-\frac{2}{3}\right|\\
& =\left|-\frac{4}{3}\right|\\
& =\frac{4}{3}\textrm{ square units}
\end{align*}\]
Given the function \(f(x)=x^{2}-1\), find the area bounded by the curve, the \(x\)-axis and the lines \(x=0\) and \(x=2\).
This is the same function as in Example 3. First, we graph the function.
As the function crosses the \(x\)-axis, between \(0\) and \(2\), we will have to evaluate two integrals. The first from \(0\) to \(1\) and the second from \(1\) to \(2\).
\[\begin{align*} \textrm{Area} & = \left|\int_{0}^{1}\left(x^{2}-1\right)dx\right|+\left|\int_{1}^{2}\left(x^{2}-1\right)dx\right|\\
& =\left|\left[\frac{1}{3}x^{3}-x\right]_{x=0}^{x=1}\right|+\left|\left[\frac{1}{3}x^{3}-x\right]_{x=1}^{x=2}\right|\\
& = \left|\frac{1}{3}-1\right|+\left|\frac{8}{3}-2-\left(\frac{1}{3}-1\right)\right|\\
& = \left|-\frac{2}{3}\right|+\left|\frac{8}{3}-\frac{6}{3}-\left(-\frac{2}{3}\right)\right|\\
& = \frac{2}{3}+\frac{2}{3}+\frac{2}{3}\\
& = 2\textrm{ square units}
\end{align*}\]
Exercise – finding the area under the curve
Find the area enclosed by the curve \(y=6x-x^{2}\) and the \(x\)-axis.
Find the area bounded by the curve \(y=\sin(x)\), the \(x\)– axis and the line \(x=\dfrac{\pi}{2}\).
Find the area bounded by the curve \(y=x^{2}-3x+2\) and the \(x\)-axis between \(x=0\) and \(x=2\).
