Find the average value of the function f(x) = 5x^3 on the interval [-2,2] and de

Question

Find the average value of the function f(x) = 5x^3 on the interval [-2,2] and determine a number c in this interval for which f(c) is equal to the average value. a) Average value = 5, c = 1 b) Average value = 0, c = 5 c) Average value = 80, c = -1 d) Average value = 0, c = 0 e) Average value = 80, c = 0 Question 2 Find the average value of the function f(x) = |4x| on the interval [-6,6] and determine a number C in this interval for which f(x) is equal to the average value. a) Average value = 0, c = +/-3 b) Average value = 24, c = -3 c) Average value = 12, c = +/-3 d) Average value =0, c = 2 e) Average value 12, c = +/-2

Explanation / Answer

f(x) = 5x^3

1/(2 - (-2)) * (integr from -2 to 2) 5x^3*dx

(1/4) * 5x^4/4

(1/4)*[5(2)^4/4 - 5(-2)^4/4]

1/4*0

0

Average = 0

5c^3 = 0
c^3 = 0
c = cuberoot(0)
c = 0

So, option D

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(1 / 0 - (-6)) * int (-4x) + (1 / (6 - 0) * integral of 4x

(1/6)*(-2x^2) + (1/6)*(2x^2)

Plug in limits :

1/6[-2(0)^2 + 2(-6)^2] + 1/6[2(6)^2 - 2(0)^2]

6 + 6

12

|4c| = 12
4c = 12 and 4c = -12
c = 3 and -3

So, option C

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