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Find The Average Value Of The Following Function On The Given Interval F X =cos X [-

Find the average value of the function on the given interval: f(x)=cos12x, [0,pi/2]?

The average value of the expression is the slope of the secant formula....

m = (f(b) - f(a))/(b - a) at [a,b]

Here, by using that formula, we have..

m = (cos(12 * π/2) - cos(12 * 0))/(π/2 - 0)
= (cos(6π) - cos(0))/(π/2)
= (1 - 1)/(π/2)
= 0

I hope this helps!

Find the average value of the following function on the interval [-pi,pi ].?

Integrate from -pi to pi. Then multiply that integral by 1 / ( pi - -pi) = 1 / 2pi.

Find the average value have of the function h on the given interval. h(x) = 3 cos^4 x sin x, [0, π]?

(1/π) ∫3cos⁴x sinx dx, from x = 0 to π
= (3/π) ∫cos⁴x sinx dx
Let u = cosx, then du = -sinx dx. The boundaries x = 0 to π become u = 1 to -1. In terms of u, the integral becomes:
(-3/π) ∫u⁴ du, from u = 1 to -1
= -3/(5π) (u⁵), from u = 1 to -1
= -3/(5π) [-1 - 1] = -3/(5π) (-2) = 6/(5π)

The average value of the function f(x)=cos(x/2) on the closed interval[-4,0] is what?

∫cos(x/2)dx
= 2sin(x/2)

Apply the limits of [-4,0] to get the area

2sin(0) - 2sin(-2)
= 0 + 2sin(2)
= 2sin(2)

Mean value
= area/span
= 2sin(2)/(0-(-4))
= 0.5sin(2)

E

Find the average of the function over the given interval?

All values of x in the interval for which the function equals its average value. (Round your answer to three decimal places.) f(x)=2cos(x), [0, pi/2]

anwser has both (x,y) ( ) in webassign

Find the average value of f(x) = (cos^4 x)(sinx) on the interval [0,2]?

I still need help sometimes, too. So I like to answer questions like yours to keep my math fresh and fun.

I like this one because it is an integration by substitution and set up to be easy.

substitute u = cos(x) and du = -sin(x) dx
= - ∫ u^4 du by substitution
and ∫ u^4 du is u^5/5
= - u^5/5
Substitute back for u = cos(x):
= -1/5 cos^5(x)
and evaluate on [0,2]
= [-1/5 cos^5(2)] - [-1/5 cos^5(0)]
and simple math now

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If the function of f given is f(x) = 3 In (2+x^2) cos x, how do you find the average value of f on the closed interval [2,6]?

You take the integral of the function with the given limits and divide by the width.Problem is, the integral is not so easy. I would use wolfram alpha or numeric methods.An approximation of the value is (f(6)+f(2))/2, so you just take the average of the two limit values (or approximate the integral with a rectangle of this height)According to Wolfram|Alpha, the answer is -0.557612.If you ask it for an indefinite integral, you get a very crazy function: Wolfram|Alpha

Find the average value of : f(x) = 4sinx + 5cosx on the interval [0,19pi/6 ]?

I have created a graph of this function for you. To see it click on the following link:

http://i369.photobucket.com/albums/oo133/gerryrains/4timessinxminus5timescosx.jpg

As you can see there is going to be massive cancellation of negative and positive parts of the curve.

The formula for the average value is:

19π/6
.....∫ (4 * sin(x)) - (5 * cos(x)) dx
....0
--------------------------------------...
...............19π/6

I'll just concentrate on the integral in the numerator for now.

19π/6
.....∫ (4 * sin(x)) - (5 * cos(x)) dx = (-4 * cos(x)) + (5 * sin(x)) from 0 to 19π/6
....0

sin(0) = 0
cos(0) = 1
sin(19π/6) = sin(7π/6) = -sin(π/6) = -(1/2)
cos(19π/6) = cos(7π/6) = -cos(π/6) = -((sqrt(3)) / 2)

So the average value is:

[((2) * sqrt(3)) - (5/2) + 4 - 0] / (19π/6) = approximately

(3.46410 + (3/2) = (4.96410 / 9.94838) = approximately

0.49899.....<<<<<.....Answer

Note: since π is an irrational number this is all approximate although it is certainly accurate to 5 decimal places.
.

How can I find the minimum value for the following function?

[math]f(x) = 2.2(sinx)^2 + (2cosx)^2[/math] [math]=[/math] [math]2(1 - cos(2x)) + (1 + cos(2x))^2[/math] [math]= 2 - 2cos(2x) + 1 + 2cos(2x) + cos^2(2x) [/math] [math]= cos^2(2x) + 3[/math][math]cos^2(2x)є[0,1][/math] thus [math]f(x)є[3,4][/math]Thus minimus value is 3 and max value is 4, the answer to your question 9 and 10. Don’t directly go for maxima or minima if u are able to manipulate the trigonometric functions.

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