What is the average value of cos(x) on the interval [-2,7] ? Choose 1 answer: (A) (sin(7)+sin(2))/(9) (B) (sin(7)-sin(2))/(9) (C) (sin(7)+sin(2))/(5) (D) (sin(7)-sin(2))/(5)

Formula Application: To find the average value of a continuous function like cos(x) over an interval [a, b], we use the formula: Average value = (1/(b-a)) * ∫ from a to b of f(x) dx Here, f(x) = cos(x), a = -2, and b = 7.Interval Width Calculation: First, we calculate the width of the interval, which is b - a. Width of interval = 7 - (-2) = 7 + 2 = 9Integration of cos(x): Now, we need to integrate cos(x) from -2 to 7. The integral of cos(x) is sin(x), so we evaluate sin(x) from -2 to 7. ∫ from -2 to 7 of cos(x) dx = sin(7) - sin(-2)Evaluation of Integral: Since sin(-x) = -sin(x), we can rewrite sin(-2) as -sin(2). So, ∫ from -2 to 7 of cos(x) dx = sin(7) - (-sin(2)) = sin(7) + sin(2)Average Value Calculation: Now, we divide the result of the integration by the width of the interval to find the average value. Average value = (1/9) * (sin(7) + sin(2))Matching Answer Choice: We look at the given answer choices to find the one that matches our result. The correct answer choice that matches our result is (A) (sin(7)+sin(2))/(9).

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What is the average value of
cos(x) on the interval
[-2,7] ?
Choose 1 answer:
(A)
(sin(7)+sin(2))/(9)
(B)
(sin(7)-sin(2))/(9)
(C)
(sin(7)+sin(2))/(5)
(D)
(sin(7)-sin(2))/(5)

What is the average value of $\cos (x)$ on the interval $[-2,7]$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $\frac{\sin (7)+\sin (2)}{9}$ $\newline$ (B) $\frac{\sin (7)-\sin (2)}{9}$ $\newline$ (C) $\frac{\sin (7)+\sin (2)}{5}$ $\newline$ (D) $\frac{\sin (7)-\sin (2)}{5}$

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Full solution

Q. What is the average value of

\cos (x)

on the interval

[-2,7]

\newline

Choose

1

answer:

\newline

(A)

\frac{\sin (7)+\sin (2)}{9}

\newline

(B)

\frac{\sin (7)-\sin (2)}{9}

\newline

(C)

\frac{\sin (7)+\sin (2)}{5}

\newline

(D)

\frac{\sin (7)-\sin (2)}{5}

Formula Application: To find the average value of a continuous function like $\cos(x)$ over an interval $[a, b]$ , we use the formula: $\newline$ Average value = $\frac{1}{(b-a)} \cdot \int_{a}^{b} f(x) \, dx$ $\newline$ Here, $f(x) = \cos(x)$ , $a = -2$ , and $b = 7$ .
Interval Width Calculation: First, we calculate the width of the interval, which is $b - a$ . $\newline$ Width of interval = $7 - (-2) = 7 + 2 = 9$
Integration of $\cos(x)$ : Now, we need to integrate $\cos(x)$ from $-2$ to $7$ . The integral of $\cos(x)$ is $\sin(x)$ , so we evaluate $\sin(x)$ from $-2$ to $7$ . $\int_{-2}^{7} \cos(x) \, dx = \sin(7) - \sin(-2)$
Evaluation of Integral: Since $\sin(-x) = -\sin(x)$ , we can rewrite $\sin(-2)$ as $-\sin(2)$ . So, $\int_{-2}^{7} \cos(x) \, dx = \sin(7) - (-\sin(2)) = \sin(7) + \sin(2)$
Average Value Calculation: Now, we divide the result of the integration by the width of the interval to find the average value. $\newline$ Average value = $(\frac{1}{9}) * (\sin(7) + \sin(2))$
Matching Answer Choice: We look at the given answer choices to find the one that matches our result. $\newline$ The correct answer choice that matches our result is (A) $(\sin(7)+\sin(2))/(9)$ .