Question

Let f be the function shown below, with domain the closed interval [0, 6]. The graph of f shown below, composed of two semicircles...

Answer 1

@saifoo.khan could you help me?

Answer 2

Sorry. :l

Answer 3

@mertsj

Answer 4

To answer A, I used the area formula for a circle and took the half of it, and subtracted the smaller semicircle from the large due to "negative area". I got 9.428

Answer 5

@myininaya This is a question for you or Satellite or Turing

Answer 6

isn't the answer to b):
f(x) = +sqrt( (x - 1)^2 - 2) and since h' = f -> f(5/2) = 1/2?

Answer 7

@gordonj005 How did you come up with that equation?

Answer 8

it's just a rewritten equation of a circle, so:
for x in [0, 4] -> (x-2)^2 + y^2 = 4 (radius 2, shifted 2 left, my equation above is off by 2)
for x in [4, 6] -> (x - 5)^2 + y^2 = 1 (radius 1, shifted 5 left)

so the function is y = {+sqrt(4 - (x-2)^2) for x in [0, 4], -sqrt( 1 - (x - 5)^2) for x in [4, 6]}
by the fundamental theorem of calculus, since h is defined as int[0 -> x] y then h' is just y. so evaluating h'(5/2) by our peicewise function, this is simply:
+sqrt( 4 - (5/2 - 2)^2) = +sqrt(4 - 1/4) = +sqrt(3*5/4) = (1/2)(sqrt3)(sqrt5)
(sorry for the late reply)