Question

I've got 1/2 of this one just not sure what I'm missing

Answer 1

Let
F(x)∫0xsin(2t)dt
. Sketch a graph of y = sin(2t) and use it to answer the questions.
F(Π)=??=0
On the interval [0,pi], for what values of x is F(x) positive? If F(x) is never positive on this interval, enter none in both boxes.
0< x <?
On the interval [0,\pi], for what values of x is F(x) negative? If F(x) is never negative on this interval, enter none in both boxes.
?< x < ?

Answer 2

sorry messed it up
\[F(x)\int\limits_{0}^{x}\sin (2t)dt\]

Answer 3

negative is none and I thought the 2nd value should be pi/2 but it isn't

Answer 4

\[F(\pi)=\int_0^x\sin(2t)dt=\left[\frac{\cos(2t)}{2}\right]_0^x={1\over2}[\cos(2x)-\cos(0)]=\frac{\cos(2x)-1}{2}\]
Attached is F(x) plot

Answer 5

Ok, so this is the only part I still need and I don't follow what it wants for the range apparently

On the interval [0,pi], for what values of x is F(x) positive? If F(x) is never positive on this interval, enter none in both boxes. 0< x <?

Answer 6

F(x) +ve: none.
F(x) -ve: entire domain excluding 0 and pi at which it is "0" (neither positive nor negative)

Answer 7

0<x<pi

Answer 8

yeah I just got that apparently I just needed to leave it and go back to it