Areas with trigonometric functions

Finding areas between trigonometric curves and the x-axis, splitting at the x-intercepts, and finding areas between two trigonometric curves.

Areas and the $x$-axis

Example 1●●

Find the area between $y = \sin x$ and the $x$-axis from $x = 0$ to $x = 2\pi$.

The curve is above the axis on $[0, \pi]$ and below it on $[\pi, 2\pi]$.

$\displaystyle\int_0^{\pi} \sin x\,dx = 2$ and $\displaystyle\int_{\pi}^{2\pi} \sin x\,dx = -2$.

Area $= 2 + 2 = 4$ square units. (The integral over $[0, 2\pi]$ is $0$.)

Between two curves

Between y = sin x and y = cos x

The curves meet at x = π/4 and x = 5π/4. Between them, sine is the upper curve.

Example 2●●●

Find the area enclosed between $y = \sin x$ and $y = \cos x$ from $x = \dfrac{\pi}{4}$ to $x = \dfrac{5\pi}{4}$.

$\displaystyle\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} (\sin x - \cos x)\,dx = \Big[-\cos x - \sin x\Big]_{\frac{\pi}{4}}^{\frac{5\pi}{4}}$

$= \left(\tfrac{1}{\sqrt2} + \tfrac{1}{\sqrt2}\right) - \left(-\tfrac{1}{\sqrt2} - \tfrac{1}{\sqrt2}\right) = \tfrac{4}{\sqrt2} = 2\sqrt2$ square units.

Example 3●

Find the area under $y = \sec^2 x$ from $x = 0$ to $x = \dfrac{\pi}{3}$.

$\Big[\tan x\Big]_0^{\frac{\pi}{3}} = \sqrt3$ square units.

Flashcards

Area under one arch of $y = sin x$

$2$

$displaystyleint_0^{2pi} sin x,dx$

$0$: the two arches cancel.

Where $sin x = cos x$

$x = \tfrac{\pi}{4}, \tfrac{5\pi}{4}, \dots$

Practice

Find the area between $y = \sin x$ and the $x$-axis from $x = 0$ to $x = 2\pi$.
Evaluate $\displaystyle\int_0^{2\pi} \sin x\,dx$.
Find the area between $y = \sin x$ and $y = \cos x$ from $x = \dfrac{\pi}{4}$ to $x = \dfrac{5\pi}{4}$. Type $\sqrt2$ as sqrt2.
Find the area under $y = \cos x$ from $x = 0$ to $x = \dfrac{\pi}{2}$.
Find the area under $y = \sec^2 x$ from $x = 0$ to $x = \dfrac{\pi}{3}$.
Find the area under $y = \sin 2x$ from $x = 0$ to $x = \dfrac{\pi}{2}$.
Find the area under $y = 3\sin x$ from $x = 0$ to $x = \pi$.
Find the area between $y = \cos x$ and the $x$-axis from $x = 0$ to $x = \pi$.
Why is the area between $y = \sin x$ and the axis on $[0, 2\pi]$ not equal to $\displaystyle\int_0^{2\pi} \sin x\,dx$?