Calculus / Unit 9: Applications of Integrals
9.2 Intersections and Splitting Regions
Notes
When no limits are given, find where the curves intersect by setting them equal.
- Example: $y = x^2$ and $y = 2x$ meet where $x^2 = 2x$, so $x = 0$ and $x = 2$. The line is on top, so area $\displaystyle = \int_0^2 (2x - x^2)\,dx = 4 - \frac{8}{3} = \frac{4}{3}$.
- If the curves cross inside the interval, split at the crossing and switch which one is on top.
- A plain integral can cancel positive and negative parts, but area never cancels.
Practice
Find the area enclosed by $y = x^2$ and $y = 2x$.
They meet at $x = 0$ and $x = 2$. $\displaystyle \int_0^2 (2x - x^2)\,dx = \frac{4}{3}$.
Find the total area between $y = x$ and $y = x^3$ on $[-1, 1]$.
The curves cross at $-1$, $0$, $1$. By symmetry, area $\displaystyle = 2\int_0^1 (x - x^3)\,dx = 2\left(\frac{1}{2} - \frac{1}{4}\right) = \frac{1}{2}$.
What is the total area between $y = \sin x$ and the $x$-axis on $[0, 2\pi]$?
$\displaystyle \int_0^{2\pi} \sin x\,dx = 0$ because parts cancel, but area is $\displaystyle 2\int_0^\pi \sin x\,dx = 4$.
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