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Section 4.7 Area under curves (IN7)

Remark 4.7.1.

A geometrical interpretation of

\begin{equation*} \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i) \Delta x = \int_a^bf(x)dx \end{equation*}

(Definition 4.5.3) defines \(\int_a^bf(x)dx\) as the net area between the graph of \(y=f(x)\) and the \(x\)-axis. By net area, we mean the area above the \(x\)-axis (when \(f(x)\) is positive) minus the area below the \(x\)-axis (when \(f(x)\) is negative).

As the number of subdivisions increases, the Riemann sum more closely appears to measure the net area between a curve and the x-axis.
Figure 71. Improving approximations of \(\int_0^5(x-2)(x-4)dx\)

Activity 4.7.2.

(a)

Write the net area between \(f(x)=6 \, x^{2} - 18 \, x\) and the \(x\)-axis from \(x=2\) to \(x=7\) as a definite integral.

(b)

Evaluate this definite integral to verify the net area is equal to \(265\) square units.

Observation 4.7.3.

In order to find the total area between a curve and the \(x\)-axis, one must break up the definite integral at points where \(f(x)=0\text{,}\) that is, wherever \(f(x)\) may change from positive to negative, or vice versa.

The total area is illustrated by breaking up the integral from 0 to 5 at 2 and 4 where (x-2) and (x-4) are equal to 0.
Figure 72. Partitioning \(\int_0^5(x-2)(x-4)dx\) at \(x=2\) and \(x=4\text{.}\)

Since \(f(x)=(x-2)(x-4)\) is zero when \(x=2\) and \(x=4\text{,}\) we may compute the total area between \(y=(x-2)(x-4)\) and the \(x\)-axis using absolute values as follows:

\begin{equation*} \text{Area} = {\color{blue} \left|\int_0^2(x-2)(x-4)dx\right|}+ {\color{red} \left|\int_2^4(x-2)(x-4)dx\right|}+{\color{blue}\left|\int_4^5(x-2)(x-4)dx\right|} \end{equation*}

Activity 4.7.4.

Follow these steps to find the total area between \(f(x)=6 \, x^{2} - 18 \, x\) and the \(x\)-axis from \(x=2\) to \(x=7\text{.}\)

(a)

Find all values for \(x\) where \(f(x)=6 \, x^{2} - 18 \, x\) is equal to \(0\text{.}\)

(b)

Only one such value is between \(x=2\) and \(x=7\text{.}\) Use this value to fill in the \(\unknown\) below, then verify that its value is \(279\) square units.

\begin{equation*} \text{Area} = \left| \int_{ 2 }^{ \unknown } \left( 6 \, x^{2} - 18 \, x \right) dx \right| + \left| \int_{ \unknown }^{ 7 } \left( 6 \, x^{2} - 18 \, x \right) dx \right| \end{equation*}

Activity 4.7.5.

Answer the following questions concerning \(f(x)=6 \, x^{2} - 96\text{.}\)

(a)

What is the total area between \(f(x)=6 \, x^{2} - 96\) and the \(x\)-axis from \(x=-1\) to \(x=9\text{?}\)

(b)

What is the net area between \(f(x)=6 \, x^{2} - 96\) and the \(x\)-axis from \(x=-1\) to \(x=9\text{?}\)