Question

Find the value of (0) when f '(x)=sin2t, f(π/2)=3

Answer 1

the value of 0?

Answer 2

Find the value of f(0) when f '(x)=sin2t, f(π/2)=3

Answer 3

big-O lol

Answer 4

antiderive sin(2t), or is that sin^2(t) ?

Answer 5

sin(2t)

Answer 6

sin(2t) we should know comes from at least a cos(2t) so what are we missing?

Answer 7

\[F(x)=-\frac{1}{2}\cos(2x)+C\] and
\[
f(\frac{\pi}{2})=-\frac{1}{2}\cos(\frac{\pi}{2})+C=3\]

Answer 8

solve for C, then plug in 0 and see what you get

Answer 9

Dt cos(2t) = -2 sin(2t)\[\]
Dt -cos(2t) = 2 sin(2t)\[\]
so we need to place a catch for that 2 that pops out\[\]
-cos(2t)/2 seems to work; +C of course

Answer 10

satellite did you use the u-substitution

Answer 11

there is no magic to a "u-sub" .... its just a device to see the work easier

Answer 12

so how did he or she come up with -1/2cos2x?

Answer 13

its called; this is a simple enough antiderivative to be able to predict where it came from

Answer 14

there is no "u-sub" magic :)

Answer 15

if you want, you can work it all out in a u-sub

Answer 16

antiD sin(2t) dt; u = 2t; du = 2 dt; du/2 = dt\[\]
antiD sin(u) du/2 = -cos(u)/2; -cos(2t)/2

Answer 17

but if you can see where the function came from to begin with and understand the mechanics that got it there; the rest of that is superflous