Area under a curve and between curves

Integration is used to determine the area under a curve and between curves. The method is very similar to solving definite integrals. It is important to note whether an area is above or below the \(x\)-axis, as this will determine the sign, positive or negative, of the area.

Use this page to revise the following concepts within area under a curve and between curves:

Signed area

The signed area refers to the net area calculated by a definite integral. That is, an algebraic sum of areas rather than the total area.

Let \(f\left(x\right)\) be a continuous function over \(x \in [a,b]\):

\(\int_{b}^{a}f\left(x\right) dx\) gives the signed area bounded between \(f(x)\) and the \(x\) axis between \(x = a\) and \(x = b\).

Hence, when using integration to find areas bound by functions it is essential to consider whether a function is above, below, or crosses the \(x\)-axis when determining a suitable integral to find the area.

  • Areas above the \(x\)-axis are positive.
  • Areas below the \(x\)-axis are negative.

When finding the signed area, subtract the areas that are below the axis from the areas that are above it.

Worked Example

Find the area bounded between the line \(y = 2x + 4\) and the \(x\)-axis between \(x = -4\) and \(x = 2\):

A graph of the linear function f(x) = x, showing the area between the curve and the x-axis over the interval [-4, 4]. The region from x = -4 to 0 is shaded in blue and labelled A₁, representing negative area. The region from x = 0 to 4 is shaded in orange and labelled A₂, representing positive area. The triangle A₁ has a height of 4 and base of 4, and the triangle A₂ has height 8 and base 4. This illustrates signed area in definite integrals.>

We can use triangles to find the area:

\[\begin{align}A_{1} &= \frac{1}{2} \times2 \times 4 &= 4 \\ A_{2} &= \frac{1}{2} \times 4 \times 8 &=16\end{align}\]

The total area is \(A_1 + A_2 = 20\)
However, the signed area, and thus the definite integral, is \(- A_1 + A_2 = 12\)

Area under a curve

Recall that \(\int_{a}^{b}f\left(x\right)dx\) gives the signed area under a curve.

To find the total area, the curve must be examined to determine if it lies above, below or crosses the \(x\)-axis.

Using symmetry properties of the curve being integrated can reduce the complexity of the integral required to be solved. For instance, the area bounded between the \(x\)-axis and the curve \(y = 1 - x^{2}\) is given but the integral \(\int_{-1}^{1}1 - x^{2}dx\). However, as this curve is symmetric about the \(y\)-axis, this area can also be found by calculating half the area and then doubling it, the corresponding integral would be \(2 \times \int_{0}^{1}1 - x^{2}dx\)

Worked Example

Example 1

Determine the area bounded by the function \(f\left(x\right) = \left(x - 3\right)^2\) and the \(x\)-axis between \(x = 0\) and \(x = 3\).

\(f(x)\) is above the \(x\)-axis so the area can be found directly:

\[\begin{align}&\int_{0}^{3}\left(x - 3\right)^{2}dx \\ =&[\frac{1}{3}\left(x - 3\right)^{3}]^{3} \\ =&\frac{1}{3}\left(3 - 3\right)^{3} - \frac{1}{3}\left(0 - 3\right)^{3} \\ =& 0 - (-9) \\ =& 9\ \textsf{square units}\end{align}\]

Example 2

Determine the area bounded by \(y = \sin(3x)\) and the \(x\)-axis between \(x = 0\) and \(x = \pi\).

A graph of the function y = sin(3x), showing three full oscillations. Between from 0 to π, areas under the curve above the x-axis are shaded in orange, and the area below the x-axis is shaded in blue. This illustrates the concept of signed area under a periodic function over an interval.

\[\int_{0}^{\frac{\pi}{3}}\sin\left(3x\right)dx - \int_{\frac{\pi}{3}}^{\frac{2\pi}{3}}\sin\left(3x\right)dx + \int_{\frac{2\pi}{3}}^{\pi}\sin\left(3x\right)dx\]

Note that between \(x = 0\) and \(x = \pi\), this function completes exactly one and a half cycles. This symmetry allows us to focus only on the first half cycle where the function is positive, and multiply this by \(3\). The integral simplifies to:

\[\begin{align}&3\int_{0}^{\frac{\pi}{3}}\sin\left(3x\right)dx \\=& 3[-\frac{1}{3}\cos\left(3x\right)]_{0}^{\frac{\pi}{3}} \\ =&\cos\left(0\right) - \cos\left(\pi\right) \\ =& 1 - (-1) \\ =& 2\ \textsf{square units}\end{align}\]

Example 3

Determine the area bounded between \(y = e^x - 2\) and the \(x\)-axis between \(x = 0\) and \(x = \log_e{3}\).

The graph cuts the \(x\)-axis at \(x = \log_e{2}\):

A graph of the exponential function y = eˣ − 2, showing the area between the curve and the x-axis over the interval [0, ln(3)]. The region from x = 0 to ln(2), where the function is negative, is shaded in blue, and the region from x = ln(2) to ln(3), where the function is positive, is shaded in orange. This highlights the concept of signed area in definite integrals for a function with exponential growth.

Since the area between \(x = 0\) and \(x = \log_{e}2\) lies below the \(x\)-axis, the definite integral over this interval gives a negative value. To represent the area as a positive quantity, a negative sign is placed in front of the corresponding definite integral.

So, the total area will be:

\[\begin{align}& -\int_{0}^{\log_{e}2}\left(e^{x} - 2\right)dx + \int_{\log_{e}2}^{\log_{e}3}\left(e^{x} - 2\right)dx \\ =&-\left[e^{x} - 2x\right]_{0}^{\log_{e}2} + \left[e^{x} - 2x\right]_{\log_{e}2}^{\log_{e}3} \\ =&\left(\left(e^{\log_{e}2} - 2\log_{e}2\right) - \left(e^{0} - 2\left(0\right)\right)\right) + \left(\left(e^{\log_{e}3} - 2\log_{e}3\right) - \left(e^{\log_{e}2} - 2\log_{e}2\right)\right) \\ =& -\left(\left(2 - 2\log_{e}2\right) - \left(1 - 0\right)\right) + \left(\left(3 - 2\log_{e}3\right) - \left(2 - 2\log_{e}2\right)\right) \\ =& \left(2\log_{e}2 - 1\right) + \left(2\log_{e}2 - 2\log_{e}3 + 1\right) \\ =& 4\log_{e}2 - 2\log_{e}3 \\ =&2\log_{e}\left(\frac{4}{3}\right)\ \textsf{square units}\end{align}\]

Note

When providing a final answer, where possible use the units from the context of the question, e.g. \(m^2\). If no units are given in the question, it is good practice is to state the answer with ‘square units’.

Area between two curves

The concept integrals to determine areas under curves can also be used to determine the area between curves.

Consider the area between \(y = x^2\) and \(y = x\):

A graph of the functions y = x² (blue parabola) and y = x (green line), showing the region bounded between them shaded in orange. The shaded area lies between the curves from x = 0 to x = 1, where the line y = x is above the parabola y = x². This illustrates the concept of the area between two curves.>

This area can be found by finding the area under \(y = x^2\) and subtracting it from the area under \(y = x\).

\[\int_{0}^{1}x dx - \int_{0}^{1}x^{2} dx\]

This is equivalent to:

\[\int_{0}^{1}x - x^{2}dx\]

In general, the area bounded between two curves is given by:

\[\int_{a}^{b}f\left(x\right) - g\left(x\right)dx\]

For two functions \(f(x)\) and \(g\left(x\right)\) where \(f\left(x\right) \geq  g\left(x\right)\) over the interval \(x \in [a,b]\)

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Worked Example

Area bound between \(y = \sin(x)\) and \(y = \cos(x)\) from \(x \in [-\frac{7\pi}{4},\frac{5\pi}{4}]\)

  • Visualise which curve and determine which function is of higher value, and where the points of intersection between the curves occur

A graph of y = sin(x) (blue) and y = cos(x) (green) with the region between the two curves shaded in orange from x = -7π/4 to 5π/4. The shaded area represents the area between the two circular functions over the interval. Within this interval the curves intersect at (−7π⁄4, √2⁄2), (−3π⁄4, −√2⁄2), (π⁄4, √2⁄2) and (5π⁄4, −√2⁄2). From (−7π⁄4, −3π⁄4) the sine function is higher, within (−3π⁄4, π⁄4) the cos function is higher, and (π⁄4, 5π⁄4) the sin function is again higher.>

  • There are 3 distinct areas between the curves in our interval of interest. We subtract the definite integral of the lower function from the higher function within each area.

\[
\text{Area} = \int_{-\frac{7\pi}{4}}^{-\frac{3\pi}{4}}  \sin(x) - \cos(x)  dx
+ \int_{-\frac{3\pi}{4}}^{\frac{\pi}{4}} \cos(x) - \sin(x)  dx
+ \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \sin(x) - \cos(x)  dx
\]

  • However, notice the symmetry; in this instance each of the three areas are equal. Therefore, we can simply solve

\[\begin{align} 3\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}}\sin\left(x\right) - \cos\left(x\right)dx &= 3\left[-\cos\left(x\right) - \sin\left(x\right)\right]_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \\ &= -3\left[\cos\left(x\right) + \sin\left(x\right)\right]_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \\ &=-3\left(\left(\cos\left(\frac{5\pi}{4}\right) + \sin\left(\frac{5\pi}{4}\right)\right) - \left(\cos\left(\frac{\pi}{4}\right) + \sin\left(\frac{\pi}{4}\right)\right)\right) \\ &=-3\left(\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}\right) - \left(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}\right)\right) \\ &=6\sqrt{2}\end{align}\]

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