The graphs of the functions $f(x)=\sin (x)$ and $g(x)=\frac{1}{2}$ intersect at $2$ points on the interval $0<x<\pi$ . $\newline$ What is the area of the region bound by the graphs of $f(x)$ and $g(x)$ between those points of intersection ? $\newline$ Choose $1$ answer: $\newline$ (A) $\frac{\pi}{3}$ $\newline$ (B) $\frac{\pi}{2}$ $\newline$ (C) $2-\frac{\pi}{2}$ $\newline$ (D) $\sqrt{3}-\frac{\pi}{3}$

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Q. The graphs of the functions

f(x)=\sin (x)

and

g(x)=\frac{1}{2}

intersect at

2

points on the interval

0<x<\pi

\newline

What is the area of the region bound by the graphs of

f(x)

and

g(x)

between those points of intersection ?

\newline

Choose

1

answer:

\newline

(A)

\frac{\pi}{3}

\newline

(B)

\frac{\pi}{2}

\newline

(C)

2-\frac{\pi}{2}

\newline

(D)

\sqrt{3}-\frac{\pi}{3}

Find Intersection Points: First, we need to find the points of intersection between $f(x) = \sin(x)$ and $g(x) = \frac{1}{2}$ on the interval $0 < x < \pi$ . To do this, we set $f(x)$ equal to $g(x)$ and solve for $x$ : $\sin(x) = \frac{1}{2}$
Calculate Area Between Curves: We know that $\sin(x) = \frac{1}{2}$ at $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$ within the interval $0 < x < \pi$ . These are the two points of intersection.
Calculate Integral of $\sin(x)$ : Next, we calculate the area of the region bound by the two graphs between $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$ . The area under $f(x)$ from $\frac{\pi}{6}$ to $\frac{5\pi}{6}$ is the integral of $\sin(x)$ from $\frac{\pi}{6}$ to $\frac{5\pi}{6}$ . The area under $g(x)$ from $\frac{\pi}{6}$ to $\frac{5\pi}{6}$ is the integral of $x = \frac{\pi}{6}$ $2$ from $\frac{\pi}{6}$ to $\frac{5\pi}{6}$ . The area between the two curves is the difference between these two areas.
Calculate Integral of $(1)/(2)$ : We calculate the integral of $\sin(x)$ from $\pi/6$ to $5\pi/6$ : $\int_{\pi/6}^{5\pi/6} \sin(x) \, dx = [-\cos(x)]_{\pi/6}^{5\pi/6}$ $= -\cos(5\pi/6) - (-\cos(\pi/6))$ $= -(-\sqrt{3}/2) - (-1/2)$ $= \sqrt{3}/2 + 1/2$
Subtract Areas: We calculate the integral of $(1)/(2)$ from $\pi/6$ to $5\pi/6$ : $\int_{\pi/6}^{5\pi/6} (1)/(2) \, dx = (1)/(2) * x$ from $\pi/6$ to $5\pi/6$ $= (1)/(2) * (5\pi/6 - \pi/6)$ $= (1)/(2) * (4\pi/6)$ $= (1)/(2) * (2\pi/3)$ $= \pi/3$
Simplify Expression: Now, we subtract the area under $g(x)$ from the area under $f(x)$ to find the area between the two curves: $\newline$ Area = $(\sqrt{3}/2 + 1/2) - (\pi/3)$ $\newline$ = $(\sqrt{3}/2 + 1/2) - (\pi/3)$ $\newline$ = $(3\sqrt{3} + 3)/(6) - (2\pi)/(6)$ $\newline$ = $(3\sqrt{3} + 3 - 2\pi)/(6)$ $\newline$ = $(3\sqrt{3} - 2\pi + 3)/(6)$
Simplify Expression: Now, we subtract the area under $g(x)$ from the area under $f(x)$ to find the area between the two curves: $\newline$ Area = $(\sqrt{3}/2 + 1/2) - (\pi/3)$ $\newline$ = $(\sqrt{3}/2 + 1/2) - (\pi/3)$ $\newline$ = $(3\sqrt{3} + 3)/(6) - (2\pi)/(6)$ $\newline$ = $(3\sqrt{3} + 3 - 2\pi)/(6)$ $\newline$ = $(3\sqrt{3} - 2\pi + 3)/(6)$ We simplify the expression to match one of the given answer choices: $\newline$ $(3\sqrt{3} - 2\pi + 3)/(6)$ $\newline$ = $(\sqrt{3} - \pi/3 + 1/2)$ $\newline$ This matches answer choice (D) when we multiply the entire expression by $2/2$ to get a common denominator for $f(x)$ $0$ and $f(x)$ $1$ : $\newline$ = $f(x)$ $2$ $\newline$ = $f(x)$ $3$