Introduction

Welcome back! Today’s topic is essentially an extension of our last two topics, as instead of having to find the area of one region only, we have to find the area of multiple and add them together. As we saw in the last few topics, we can find the area between curves by finding where they intersect, then taking the integral of the top curve minus the bottom curve. Now, we’ll be dealing with functions that include multiple points of intersection over the interval, meaning we’ll need to use multiple integrals to calculate the total area. In case you were wondering how this would be used in real life, being able to calculate the area between multiple curves allows you to find the net difference between one quantity and another quantity. Since we’ve already covered how to find the area between curves, we’re actually going to go straight into the practice section, as you should know how to add together integrals that cover different intervals.

Practice

Practice Answers

1)

First, we’ll find the intercepts so we know the bounds of our integrals.

Finding Intercepts:

To find the intercepts, we’ll set the functions equal to each other and solve for x.

$\cos (x)=\cos (3x)$

$\cos (x)-\cos (3x)=0$

$2\sin (2x)\sin (x)=0$

$x=0,\frac{\pi }{2},\pi$

Since we have three intersections, we need to find which function is greater on $\left[0,\frac{\pi }{2}\right]$ and $\left[\frac{\pi }{2},\pi \right]$.

On $\left[0,\frac{\pi }{2}\right]$, we’ll test $\frac{\pi }{4}$.

$\cos \left(\frac{\pi }{4}\right)=\frac{\sqrt{2}}{2}$

$\cos \left(3\left(\frac{\pi }{4}\right)\right)=\frac{-\sqrt{2}}{2}$

Therefore $y=cos(x)$ is the upper curve and $y=cos(3x)$ is the lower curve.

On $\left[\frac{\pi }{2},\pi \right]$, we’ll test $\frac{3\pi }{4}$.

$\cos \left(\frac{3\pi }{4}\right)=\frac{-\sqrt{2}}{2}$

$\cos \left(3\left(\frac{3\pi }{4}\right)\right)=\frac{\sqrt{2}}{2}$

Therefore $y=\cos (3x)$ is the upper curve and $y=\cos (x)$ is the lower curve.

Plugging in:

Now, we plug in our functions and solve.

$A={\int }_0^{\frac{\pi }{2}}(\cos (x)-\cos (3x))dx+{\int }_{\frac{\pi }{2}}^{\pi }(\cos (3x)-\cos (x))dx$

$A={\left[\sin (x)-\frac{\sin (3x)}{3}\right]}_0^{\frac{\pi }{2}}+{\left[\frac{\sin (3x)}{3}-\sin (x)\right]}_{\frac{\pi }{2}}^{\pi }$

$A=\left[\left(\sin \left(\frac{\pi }{2}\right)-\frac{\sin \left(3\left(\frac{\pi }{2}\right)\right)}{3}\right)-\left(\sin (0)-\frac{\sin (3(0))}{3}\right)\right]+\left[\left(\frac{\sin (3(\pi ))}{3}-\sin (\pi )\right)-\left(\frac{\sin (3\left(\frac{\pi }{2}\right))}{3}-\sin \left(\frac{\pi }{2}\right)\right)\right]$

$A=\left[\left(1-\frac{(-1)}{3}\right)-\left(0-\frac{0}{3}\right)\right]+\left[\left(\frac{0}{3}-0\right)-\left(\frac{(-1)}{3}-1\right)\right]$

$A=\left[\left(\frac{4}{3}\right)-(0)\right]+\left[(0)-\left(\frac{-4}{3}\right)\right]$

$A=\frac{4}{3}+\frac{4}{3}$

$A=\frac{8}{3}$

Therefore, the total area enclosed between the curves $y=cos(x)$ and $y=cos(3x)$ on the interval $[0,\pi ]$ is $\frac{8}{3}$.

2)

First, we’ll find the intercepts so that we can get the bounds of our integrals.

Finding Intercepts:

To find the intercepts, we’ll set the functions equal to each other and solve for x.

$\sin (x)=\sin (2x)$

$x=0,\frac{\pi }{3},\pi$ 

Since we have three intersections, we need to find which function is greater on $\left[0,\frac{\pi }{3}\right]$ and $\left[\frac{\pi }{3},\pi \right]$.

On $\left[0,\frac{\pi }{3}\right]$, we’ll test $\frac{\pi }{4}$.

$\sin \left(\frac{\pi }{4}\right)=\frac{\sqrt{2}}{2}$

$\sin \left(2\left(\frac{\pi }{4}\right)\right)=1$

Therefore $y=\sin (2x)$ is the upper curve and $y=\sin (x)$ is the lower curve.

On $\left[\frac{\pi }{3},\pi \right]$, we’ll test $\frac{\pi }{2}$.

$\sin \left(\frac{\pi }{2}\right)=1$

$\sin \left(2\left(\frac{\pi }{2}\right)\right)=0$

Therefore $y=\sin (x)$ is the upper curve and $y=\sin (2x)$ is the lower curve.

Plugging in:

Now, we plug in our functions and solve.

$A={\int }_0^{\frac{\pi }{3}}(\sin (2x)-\sin (x))dx+{\int }_{\frac{\pi }{3}}^{\pi }(\sin (x)-\sin (2x))dx$

$A={\left[-\frac{\cos (2x)}{2}+\cos (x)\right]}_0^{\frac{\pi }{3}}+{\left[-\cos (x)+\frac{\cos (2x)}{2}\right]}_{\frac{\pi }{3}}^{\pi }$

$A=\left(\left[-\frac{\cos (2\left(\frac{\pi }{3}\right))}{2}+\cos \left(\frac{\pi }{3}\right)\right]-\left[-\frac{\cos (2(0))}{2}+\cos (0)\right]\right)+\left(\left[-\cos (\pi )+\frac{\cos (2(\pi ))}{2}\right]-\left[-\frac{\cos (2(\left(\frac{\pi }{3}\right))}{2}+\cos \left(\frac{\pi }{3}\right)\right]\right)$

$A=\left(\left[-\frac{(-1)/2}{2}+\left(\frac{1}{2}\right)\right]-\left[-\frac{1}{2}+1\right]\right)+\left(\left[-(-1)-\frac{1}{2}\right]-\left[-\left(\frac{1}{2}\right)-\frac{(-1)/2}{2}\right]\right)$

$A=\left(\left[-\frac{1}{4}+\left(\frac{1}{2}\right)\right]-\left[\frac{1}{2}\right]\right)+\left(\left[1-\frac{1}{2}\right]-\left[-\frac{1}{2}+\frac{1}{4}\right]\right)$

$A=\left(-\frac{1}{4}\right)+\left(\left[\frac{1}{2}\right]-\left[-\frac{1}{4}\right]\right)$

$A=-\frac{1}{4}+\frac{3}{4}$

$A=\frac{1}{2}$

Therefore, the total area enclosed by the curves $y=\sin (x)$ and $y=\sin (2x)$ on the interval $[0,\pi ]$ is $\frac{1}{2}$.