Alrighty then, lets get started.
Today during our study group, I learned average value thanks to Kaitlyn! So I guess I will explain an example of that since I finally get it.
EX: Find the average value of the function f(x)=12x-6x^2 over the interval -5 less than or equal to x less than or greater than 5.
use the formula: 1/b-a S equation given, then solve the definite integral (*S is integral)
1/5-(-5)= 1/10
1/10 (integral form -5 to 5) 12x-6x^2
1/10(12x-2x^3)
1/10(12(5)-2(5)^3)-1/10(12(-5)-2(-5)^3)
1/10(-190)-1/10(190)
-19-(-19)= -38
I do have quite a few questions though..
1. If I get (-1/64)sin(1/8) and I'm suppose to set it equal to 0 then solve..How do I solve that?
2. (like #11 on the packet) For S(3t/t^2+2)^3 I used substitution to solve. For my final answer I got -3/2(t^2+2)^2+c but all the answers have -3/4(t^2+2)^2+c
3. (like #3 on the packet) It says use a graphing utility to graph the function f(x)=12/6-x and locate the absolute extrema of the function on the interval [0,6). If you plug that into your calculator the graph is an asymptote, and I tried taking the derivative and plugging that in too but it is also an asymptote. So how do you get a max and min?
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1. You must multiply by -64 then take the trig inverse.
ReplyDelete2. You didn't account that the derivative of t^2 was 2t therefore you would need to put a 1/2 in front.