Math / Calculus

QuestionFind the area of the region enclosed by the curves $y=2 \cos x$ and $y=2 \cos 2 x$ for $0 \leq x \leq \pi$ .

Studdy Solution

STEP 1

1. We are given two functions: $y = 2 \cos x$ and $y = 2 \cos 2x$ .
2. We need to find the area of the region enclosed by these curves from $x = 0$ to $x = \pi$ .
3. The area between two curves is found by integrating the difference of the functions over the given interval.

STEP 2

1. Find the points of intersection of the curves within the interval $0 \leq x \leq \pi$ .
2. Determine the intervals where one function is above the other.
3. Set up the integral(s) to calculate the area between the curves.
4. Evaluate the integral(s) to find the total enclosed area.

STEP 3

Find the points of intersection by setting the functions equal:
$2 \cos x = 2 \cos 2x$
Simplify:
$\cos x = \cos 2x$
Using the identity $\cos 2x = 2\cos^2 x - 1$ , we have:
$\cos x = 2\cos^2 x - 1$
Rearrange to form a quadratic equation:
$2\cos^2 x - \cos x - 1 = 0$

STEP 4

Solve the quadratic equation:
Let $u = \cos x$ , then:
$2u^2 - u - 1 = 0$
Use the quadratic formula:
$u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
where $a = 2$ , $b = -1$ , $c = -1$ :
$u = \frac{1 \pm \sqrt{1 + 8}}{4}$ $u = \frac{1 \pm 3}{4}$
Thus, $u = 1$ or $u = -\frac{1}{2}$ .
Convert back to $x$ :
For $\cos x = 1$ , $x = 0$ .
For $\cos x = -\frac{1}{2}$ , $x = \frac{2\pi}{3}$ .

STEP 5

Determine which function is above the other in each interval:
1. From $x = 0$ to $x = \frac{2\pi}{3}$ , compare $y = 2 \cos x$ and $y = 2 \cos 2x$ .
2. From $x = \frac{2\pi}{3}$ to $x = \pi$ , compare $y = 2 \cos x$ and $y = 2 \cos 2x$ .

STEP 6

Evaluate the functions at a test point in each interval:
1. For $x = \frac{\pi}{3}$ , $2 \cos x = 2 \times \frac{1}{2} = 1$ and $2 \cos 2x = 2 \times (-\frac{1}{2}) = -1$ . So, $2 \cos x$ is above.
2. For $x = \pi$ , $2 \cos x = 2 \times (-1) = -2$ and $2 \cos 2x = 2 \times 1 = 2$ . So, $2 \cos 2x$ is above.

STEP 7

Set up the integral for each interval:
1. From $x = 0$ to $x = \frac{2\pi}{3}$ :
$\int_0^{\frac{2\pi}{3}} (2 \cos x - 2 \cos 2x) \, dx$
2. From $x = \frac{2\pi}{3}$ to $x = \pi$ :
$\int_{\frac{2\pi}{3}}^{\pi} (2 \cos 2x - 2 \cos x) \, dx$

STEP 8

Evaluate the integrals:
1. Evaluate $\int (2 \cos x - 2 \cos 2x) \, dx$ :
$\int 2 \cos x \, dx = 2 \sin x$ $\int -2 \cos 2x \, dx = -\sin 2x$
Evaluate from $0$ to $\frac{2\pi}{3}$ :
$\left[ 2 \sin x - \sin 2x \right]_0^{\frac{2\pi}{3}}$
2. Evaluate $\int (2 \cos 2x - 2 \cos x) \, dx$ :
$\int 2 \cos 2x \, dx = \sin 2x$ $\int -2 \cos x \, dx = -2 \sin x$
Evaluate from $\frac{2\pi}{3}$ to $\pi$ :
$\left[ \sin 2x - 2 \sin x \right]_{\frac{2\pi}{3}}^{\pi}$

STEP 9

Calculate the definite integrals:
1. From $0$ to $\frac{2\pi}{3}$ :
$\left[ 2 \sin x - \sin 2x \right]_0^{\frac{2\pi}{3}} = \left( 2 \sin \frac{2\pi}{3} - \sin \frac{4\pi}{3} \right) - \left( 2 \sin 0 - \sin 0 \right)$
$= \left( 2 \times \frac{\sqrt{3}}{2} - (-\frac{\sqrt{3}}{2}) \right)$
$= \left( \sqrt{3} + \frac{\sqrt{3}}{2} \right)$
$= \frac{3\sqrt{3}}{2}$
2. From $\frac{2\pi}{3}$ to $\pi$ :
$\left[ \sin 2x - 2 \sin x \right]_{\frac{2\pi}{3}}^{\pi} = \left( \sin 2\pi - 2 \sin \pi \right) - \left( \sin \frac{4\pi}{3} - 2 \sin \frac{2\pi}{3} \right)$
$= \left( 0 - 0 \right) - \left( -\frac{\sqrt{3}}{2} - 2 \times \frac{\sqrt{3}}{2} \right)$
$= \left( \frac{\sqrt{3}}{2} + \sqrt{3} \right)$
$= \frac{3\sqrt{3}}{2}$

STEP 10

Add the areas from both intervals to find the total area:
$\text{Total Area} = \frac{3\sqrt{3}}{2} + \frac{3\sqrt{3}}{2} = 3\sqrt{3}$
The area of the region enclosed by the curves is:
$\boxed{3\sqrt{3}}$

Was this helpful?

Get Studdy

Scan QR code with your device to get Studdy on iOS or Android

QR code

Studdy solves anything!

Download the app

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Parents Influencer program Contact Policy Terms