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Section 4.7 Area Under Curves (IN7)

Subsection 4.7.1 Activities

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*}
(DefinitionΒ 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).
Diagram Exploration Keyboard Controls
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Figure 73. 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.
Diagram Exploration Keyboard Controls
Key Action
Enter, A Activate keyboard driven exploration
B Activate menu driven exploration
Escape Leave exploration mode
Cursor down Explore next lower level
Cursor up Explore next upper level
Cursor right Explore next element on level
Cursor left Explore previous element on level
X Toggle expert mode
W Extra details if available
Space Repeat speech
M Activate step magnification
Comma Activate direct magnification
N Deactivate magnification
Z Toggle subtitles
C Cycle contrast settings
T Monochrome colours
L Toggle language (if available)
K Kill current sound
Y Stop sound output
O Start and stop sonification
P Repeat sonification output
Figure 74. 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 \(x=9\text{?}\)

Subsection 4.7.2 Videos

Figure 75. Video for IN7

Subsection 4.7.3 Exercises