Question

How would I solve this?
The average value of v(t) e^t 2cos t = over the domain [1, 4] is approximately:

F. −1.149 × 106
G. −2.840 X 105
H. 2.516 × 105
I. 1.688
J. 0.562

Answer 1

http://en.wikipedia.org/wiki/Average#Averages_of_functions

Answer 2

so I go \[1/(4-1)\int\limits_{1}^{4} e ^{t} 2cost \]?

Answer 3

That's right, not sure how to integrate that though.

Answer 4

@lalaly could you show me how to integrate this?

Answer 5

u = e^t
du = e^tdt
dv = costdt
v = sint

(e^t)(sint) - ∫ sint e^t dt

u = e^t
du = e^t dt
dv = sint
v = -cost dt

(e^t)(sint) - ∫ sint e^tdt
(e^t)(sin t) - [ (e^t)(-cos t) - ∫ -cost e^t dt ]

(e^t)(sin t) + (e^t)(cos t) - ∫ cost e^t dt
so
∫ cos(t e^t dt = (e^t)(sin t) + (e^t)(cos t) - ∫ cos(x) e^t dt

∫ cos(t e^t dt = (e^t)(sin t) + (e^t)(cos t) - ∫ cost e^t dt

2 ∫ cost e^t dt = (e^t)(sin t) + (e^t)(cost
∫ cost e^t dt = [ (e^t)(sint) + (e^t)(cos t) ] / 2

Answer 6

just plug in the limits from 1 to 4

Answer 7

yeah and multiply the integral by 1/3 lol

Answer 8

what's dv?

Answer 9

i integrated by parts ,,, so
u=e^t dv= costdt
du=e^tdt v=sint