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 ###
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.