(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).
Activity4.7.2.
(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.
Observation4.7.3.
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.
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:
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.