Find the area of the region bounded by the graphs of the equations.
\[y = (\sin(\pi x))^2 y = 0, x = 0, x = 17\]
I hate these stupid questions about area GRRR!
I believe I integrate normally tho....
yep... and you already did the \(\large \int sin^2(ax)dx \) earlier.... there should be no problem....
half angle formulas?e
or pythagorean?
but here's a big hint... \[\large \int_0^{17}sin^2(\pi\cdot x)dx=17\int_0^{1}sin^2(\pi\cdot x)dx \]
ya... half angle formula...
well, I need to head out to fireworks now, I'll ttyl,t hanks!
happy 4th man!!!! and yeah, why u doing math on JULY 4th? u should be throwing fireworks at the neighbors dogs....
@dpaInc cuz I gotta get it done.. Happy 4th to you :).
ROUND 2 LETS DO THIS!!!
\[\frac{17}{2\pi} \int\limits_{0}^{1} 1-\cos(2u)du\]
\[\frac{17}{2\pi} \int\limits_{0}^{1} x-\sin(2u) +c?\]
\[\frac{17}{2\pi} [ 1-\sin(2 \pi 1)] -[0-\sin(0) ]\]
\[\frac{17}{2\pi} [1] -[0]\]
\[\frac{17}{2\pi}?\]
Check your integration again, there's a mistake
was I correct with the 17/2pi?
and 1 -cos(2u) isnt' x - sin(2u)?
let me check again
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
your limits will chanfe
naa....
I don't like to change my limits, i just substitute back in for x and use that.
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
seems correct:D sorry :(
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?
hmm... according to the practice problem that uses 5 instead of 17 it's 5/2... They lik got rid of their pi.. :(
and why does it become 4pi........................
yeah I guess u is = 2pix...
Idk why the pi got cancelled tho...
wait, I'm gonna solve it
Go for it :D
\[17\int_0 ^1 \sin^2 (\pi x)dx\] \[17\int_0 ^1 (1-\cos (2 \pi x))/2 dx\] \[\frac{17}{2} [x-\frac{\sin {2\pi x}}{2\pi}]_0^1\] 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 \[\frac{17}{2}\]
Yeah... I guess in this case we don't take out the bottom part and throw it to the other side... stooopid...
did you understand?
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?
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
I meant now you find !!
Do you get it?
:)
I think it's supposed ot be 17/4pi and then the 2pi's cancel out? Idk big headache ;(
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