Lesson 19 Area between Curves

19.1 Lecture Content

19.2 Lecture Notes

19.2.1 Definite Integrals as Net Area

The definite integral \(\int_a^b f(x) \, dx\) represents the net area between the graph of \(f(x)\) and the \(x\)-axis over the interval \([a, b]\):

If \(f(x) \ge 0\) on \([a, b]\), the integral gives the total area above the \(x\)-axis.
If \(f(x) \le 0\) on \([a, b]\), the integral gives the negative of the area below the \(x\)-axis.
If \(f(x)\) changes sign on \([a, b]\), the integral gives the net area: area above minus area below.

Note: To find the total area, regardless of sign, use:

\[ \text{Total area} = \int_a^b |f(x)| \, dx \]

19.2.2 Simple Improper Integrals

An improper integral arises when:

The interval is infinite, e.g., \(\int_1^\infty \frac{1}{x^2} \, dx\)
The integrand is unbounded at a point in the interval, e.g., \(\int_0^1 \frac{1}{\sqrt{x}} \, dx\)

19.2.2.1 Infinite Interval Example:

\[ \int_1^\infty \frac{1}{x^2} \, dx = \lim_{b \to \infty} \int_1^b \frac{1}{x^2} \, dx = \lim_{b \to \infty} \left( -\frac{1}{x} \Big|_1^b \right) = 1 \]

19.2.2.2 Unbounded Function Example:

\[ \int_0^1 \frac{1}{\sqrt{x}} \, dx = \lim_{a \to 0^+} \int_a^1 \frac{1}{\sqrt{x}} \, dx = \lim_{a \to 0^+} (2\sqrt{x} \big|_a^1) = 2 \]

An improper integral converges if the limit exists, and diverges if the limit does not exist.

19.2.3 Area Between Two Curves

To find the area between two curves \(f(x)\) and \(g(x)\) on \([a, b]\), where \(f(x) \ge g(x)\):

\[ \text{Area} = \int_a^b [f(x) - g(x)] \, dx \]

Steps:

Find points of intersection: Solve \(f(x) = g(x)\) to find the limits of integration.
Identify top and bottom functions on the interval.
Set up and evaluate the integral:
- If necessary, split the interval where the curves switch order.
For vertical slices (functions of \(y\)), use: \[ \int_{c}^{d} [f(y) - g(y)] \, dy \]

Example:

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

Intersection points: \(x = 0\), \(x = 1\)
\(x \ge 0 \Rightarrow x \ge x^2\)
Area: \[ \int_0^1 (x - x^2) \, dx = \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_0^1 = \frac{1}{2} - \frac{1}{3} = \frac{1}{6} \]

19.3 Examples

Video: Solutions

Find the area under the curve of \(m(x) = 12 - 0.5x^2\) from \(1\) until the point at which the curve hits the \(x\)-axis.
Find the area bounded by \(y = 4\sqrt{x}\), \(y = 0\) and \(x = 4\).
Find the area bounded by the curve \(y = 11 + 6x + x^2\) and the x-axis, from \(x = -3\) to \(x = -1\).
Sketch the region enclosed by \(y = 5x\) and \(y = 5x^2\).
A sketch of the region enclosed by \(y = x\) and \(y = \frac{3.25x}{x^2 + 1}\). Find the area.

19.4 Practice Problems

Find area under \(y = e^x\) on \([1,4]\)
Find area under \(y = \sin(x)\) on \([0, \pi]\); net vs true area on \([0, 2\pi]\)
Area under \(y = 1/x^2\) on \([1, \infty]\)
Why does \(\int_1^\infty 1/x dx\) not converge?
(Challenge) For what \(p\) does \(\int_1^\infty \frac{1}{x^p} dx\) converge?
Find area between \(y = x^2\) and \(y = x^4\)
Find area between \(y = -|x|\) and \(y = x^2 - 2\)
(Challenge) Find net and true area between \(y = -x^2 + 4\) and \(y = x^2 - 2x\)

19.5 Self Assessment

Try to solve these in 20 minutes:

Area under \(y = \ln(x)\) on \([2,8]\)
Area bounded by \(y = x^3\) and the axes in the 4th quadrant
Area under \(y = e^{-x}\) on \([0,\infty]\)
Area between \(y = -|x|\) and \(y = x^2 - 2x\)

19.6 Lesson Checklist

This checklist is designed to help you keep track of what you need to work on. The main goal is to be aware of what you need to focus more attention on. Place an \(X\) in the appropriate box beside the skill below.

\[\begin{align*} &\textbf{Developing (D):} &&\textrm{You still need to work on this skill.}\\ &\textbf{Consistent (CON):} &&\textrm{You use the skill correctly most of the time.}\\ &\textbf{Competent (COM):} &&\textrm{You show mastery of the skill.} \end{align*}\]

Skill	D	CON	COM
Determine net and true area with \(x\)-axis.
Determine net and true area between curves.
Evaluate simple improper integrals.