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What is the area of the region between 2 consecutive points where the graphs of
f(x)=cos(x) and
g(x)=-cos(x)+2 intersect?

What is the area of the region between $2$ consecutive points where the graphs of $f(x)=\cos (x)$ and $g(x)=-\cos (x)+2$ intersect?

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Identify direct variation and inverse variation

Full solution

Q. What is the area of the region between

2

consecutive points where the graphs of

f(x)=\cos (x)

and

g(x)=-\cos (x)+2

intersect?

Find Intersection Points: Find the points of intersection between the two functions. $\newline$ To find the points of intersection, we set $f(x)$ equal to $g(x)$ : $\newline$ $\cos(x) = -\cos(x) + 2$
Solve for x: Solve the equation for x. $\newline$ $2\cos(x) = 2$ $\newline$ $\cos(x) = 1$ $\newline$ The cosine function equals $1$ at $x = 2n\pi$ , where $n$ is an integer. However, since we are looking for the first two consecutive points of intersection, we will consider $n = 0$ and $n = 1$ . $\newline$ $x = 0$ , $x = 2\pi$
Determine Area: Determine the area between the curves from $x = 0$ to $x = 2\pi$ . The area $A$ between two curves from $a$ to $b$ is given by the integral from $a$ to $b$ of the top function minus the bottom function: $A = \int_{a}^{b} (\text{top function} - \text{bottom function}) \, dx$ In this case, from $x = 0$ to $x = 2\pi$ , the top function is $x = 2\pi$ $0$ and the bottom function is $x = 2\pi$ $1$ . $x = 2\pi$ $2$
Simplify and Integrate: Simplify the integrand and calculate the integral. $\newline$ $A = \int_{0}^{2\pi} (-2\cos(x) + 2) \, dx$ $\newline$ $A = [-2\sin(x) + 2x]_{0}^{2\pi}$
Evaluate Antiderivative: Evaluate the antiderivative at the bounds and subtract. $\newline$ $A = [-2\sin(2\pi) + 2(2\pi)] - [-2\sin(0) + 2(0)]$ $\newline$ $A = [0 + 4\pi] - [0 + 0]$ $\newline$ $A = 4\pi$

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