Question

Find the area of the region bounded by the graphs of the equations.

Answer 1

\[y = (\sin(\pi x))^2 y = 0, x = 0, x = 17\]

Answer 2

I hate these stupid questions about area GRRR!

Answer 3

I believe I integrate normally tho....

Answer 4

yep... and you already did the \(\large \int sin^2(ax)dx \) earlier.... there should be no problem....

Answer 5

half angle formulas?e

Answer 6

or pythagorean?

Answer 7

but here's a big hint...
\[\large \int_0^{17}sin^2(\pi\cdot x)dx=17\int_0^{1}sin^2(\pi\cdot x)dx \]

Answer 8

ya... half angle formula...

Answer 9

well, I need to head out to fireworks now, I'll ttyl,t hanks!

Answer 10

happy 4th man!!!!
and yeah, why u doing math on JULY 4th? u should be throwing fireworks at the neighbors dogs....

Answer 11

@dpaInc cuz I gotta get it done.. Happy 4th to you :).

Answer 12

ROUND 2 LETS DO THIS!!!

Answer 13

\[\frac{17}{2\pi} \int\limits_{0}^{1} 1-\cos(2u)du\]

Answer 14

\[\frac{17}{2\pi} \int\limits_{0}^{1} x-\sin(2u) +c?\]

Answer 15

\[\frac{17}{2\pi} [ 1-\sin(2 \pi 1)] -[0-\sin(0) ]\]

Answer 16

\[\frac{17}{2\pi} [1] -[0]\]

Answer 17

\[\frac{17}{2\pi}?\]

Answer 18

Check your integration again, there's a mistake

Answer 19

was I correct with the 17/2pi?

Answer 20

and 1 -cos(2u) isnt' x - sin(2u)?

Answer 21

let me check again

Answer 22

I know the 2 goes down because sin^2(u) = 1-cos(2u)/2

but I used u sub for

u = pix
1/pi du = dx

Answer 23

your limits will chanfe

Answer 24

naa....

Answer 25

I don't like to change my limits, i just substitute back in for x and use that.

Answer 26

It's just how I learned originally, and I didn't hear about the change until later, which I didn't want to do, so I stuck to using x not u :P

Answer 27

seems correct:D
sorry :(

Answer 28

No, not at all, don't be :). I'm here to learn, and if I made a mistake I want to know about it...

So is 17/2pi the answer, or how does the area shizz work?

Answer 29

hmm... according to the practice problem that uses 5 instead of 17 it's 5/2... They lik got rid of their pi.. :(

Answer 30

and why does it become 4pi........................

Answer 31

yeah I guess u is = 2pix...

Answer 32

Idk why the pi got cancelled tho...

Answer 33

wait, I'm gonna solve it

Answer 34

Go for it :D

Answer 35

\[17\int_0 ^1 \sin^2 (\pi x)dx\]
\[17\int_0 ^1 (1-\cos (2 \pi x))/2 dx\]
\[\frac{17}{2} [x-\frac{\sin {2\pi x}}{2\pi}]_0^1\]
second term is zero , so it's just
\[\frac{17}{2} [x-\cancel{\frac{\sin {2\pi x}}{2\pi}}]_0^1\]
so we get
\[\frac{17}{2}\]

Answer 36

Yeah... I guess in this case we don't take out the bottom part and throw it to the other side... stooopid...

Answer 37

did you understand?

Answer 38

yup... So I guess it's situational when we throw things over to left of the integration? Or maybe just whole #'s and leave pi on the right?

Answer 39

you know if you take
\[\pi x=t\]
then
\[\pi dx=dt\]
so we get
\[\frac{17}{\pi} \int_0 ^1 \sin^ 2t dt\]
on interrogation, you;d get
\[\frac{17}{2\pi} [t-sin 2t]_0 ^1\]
you didn't change the limits,so put \( t=\pi x\)
\[\frac{17}{2\pi} [\pi x-sin 2\pi x]_0 ^1\]
Now substitute back

Answer 40

I meant now you find !!

Answer 41

Do you get it?

Answer 42

:)

Answer 43

I think it's supposed ot be 17/4pi and then the 2pi's cancel out? Idk big headache ;(