Calculating Areas Bounded by the x-Axis

Calculating Areas Bounded by the x-Axis

Suppose we have a function $f$. The function may look something like this:


And the area bounded by the function $f$ and the $x$-axis on the interval $[a, b]$ is the yellow shaded region above. We will now learn how to calculate these areas. It turns out that if we have a continuous function on the interval $[a, b]$, then we can calculate the area trapped between the curve using integral. In fact:

\begin{align} \mathbf{Area} = \int_a^b f(x) \: dx \end{align}
Note: If $f(x) < 0$ over the interval $[c, d]$. We note that $\int_c^d f(x) \: dx < 0$ as a result. If we want to calculate the area on the interval $[c, d]$, then we must take note of these points and take the absolute value of the integral. If we are trying to find the net area (area above x-axis subtracted by the area below the x-axis), then we don't take the absolute value of the integral. It is important to distinguish the different between area (which is strictly non-negative) and net area (which can be non-negative or negative). Therefore, $\mathrm{Area} = \biggr \rvert \int_c^d f(x) \: dx \biggr \rvert$, and $\mathrm{Net \: Area} = \int_c^d f(x) \: dx$

Example 1

Calculate the area of the function $f(x) = \cos x$ bounded by the x-axis on the interval $[0, \pi]$.


Recall that the function $\cos x$ has an x-intercept on the interval $[0, \pi]$ at $x = \frac{\pi}{2}$. By the CAST rule, we know that $\cos x > 0$, or rather $f(x) > 0$ on the interval $(0, \frac{\pi}{2})$, and that $\cos x < 0$ or $f(x) < 0$ on the interval $(\frac{\pi}{2}, \pi)$. So it follows that:

\begin{align} \mathbf{Area} = \int_0^\frac{\pi}{2} \cos x \: dx + \biggr \rvert \int_\frac{\pi}{2}^{\pi} \cos x \: dx \biggr \rvert \\ \mathbf{Area} = \sin x \biggr \rvert_{0}^{\pi/2} + \biggr \rvert \sin x \biggr \rvert_{\pi/2}^{\pi} \biggr \rvert \\ \mathbf{Area} = 1 + 1 \\ \mathbf{Area} = 2 \end{align}

Therefore the area under $f$ on the interval $[0, \pi]$ is $2$.

Note: If we wanted to calculate the net area over the interval $[0, \pi]$, then $\int_0^{\pi} \cos x \: dx = 0$.

Example 2

Calculate the area of the function $f(x) = 3x^3 - 4x^2$ over the interval $[1, 2]$.

We first need to locate any $x$-intercepts. We note that $3x^3 - 4x^2 = x^2(3x - 4)$. Therefore, $x = 0$ and $x = \frac{4}{3}$ are the x-intercepts.

We now want to see whether $f$ is positive of negative over the interval $[1, \frac{4}{3} ]$. We test $f(1.2) = -0.576 < 0$. Therefore, we get that:

\begin{align} \mathrm{Area} = \biggr \rvert \int_{1}^{4/3} 3x^3 - 4x^2 \: dx \biggr \rvert + \int_{4/3}^{2} 3x^3 - 4x^2 \: dx \\ \mathrm{Area} = \mathrm{abs} \left ( \frac{3}{4}x^4 - \frac{4}{3} x^3 \biggr \rvert_{1}^{4/3} \right ) + \frac{3}{4}x^4 - \frac{4}{3} x^3 \biggr \rvert_{4/3}^2 \\ \mathrm{Area} = \mathrm{abs} (-0.2068) + 2.1235 \\ \mathrm{Area} = 2.33 \end{align}
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