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anonymous
 5 years ago
Integrate: xsin(pix^2)dx I'm going U sub with u equaling to pix^2 but I get stuck trying to compensate for the du = 2pixdx
then dx = du/(2pix)
anonymous
 5 years ago
Integrate: xsin(pix^2)dx I'm going U sub with u equaling to pix^2 but I get stuck trying to compensate for the du = 2pixdx then dx = du/(2pix)

This Question is Closed

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0u = pi x^2 du = 2pi x dx dx = du/2pi x

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0x sin(u) sin(u)  =  du 2pi x 2pi

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so i can pull out 1/2pi in front

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0yep; it aint gonna change any :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if thats the case then the answer he has listed is wrong, the bounds are from 0 to 1 his answer listed is 1

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0you need to have a 2pi up top to counter your integration

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So when I set something dx = blah blah blah, i have to account for what im plugging in

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for instance dx = 2xdu, i need to replace dx with the 2xdu but also put in a 1/2x?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0whatch.... D(cos(2pix^2)) = 2pix sin(2pix^2) right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol newton his answers are rarely ever what they seem to be, its also a double integration though

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0post the entire questionthen; its rather difficult to step in half way and try to determine where something messed up at

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.04pi x sin(2pix^2)  dx is doable right? 4pi

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.00to1cube root y to 1 (sin(pix^2))/x^2 dxdy

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hard to type a double integral with this crappy window!

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0your u was ok but we messed up the derivative of it lol u = 2pix^2 du = 4pi x dx dx = du/4pi x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0uh why is u = 2pi...?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\iint \] \.[\iint \]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.00<y<1; and cbrt(y)<x<1 is your bounds right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0you think changeing them would make it easier?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0{SS} [ sin(pix^2)/x^2] dy dx

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0y=1 y=0 x=1 x=cbrt(y) can be changed to make it easier to start with a dy

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Amistre, brah, one day you're gonna have to give in and learn some LaTeX.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0we can use this to switch to dydx

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats what i did, makes the first one extremely easy

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0u just ad a y t hen its all good

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.00 < y < x^3 0 < x < 1 right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ya then u do the first one, and its just the regular eq with a y added to it

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i gotta work thru it to see where se go tho :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then u sub it all in and do the bounds and the x's cancel til u get xsin(pix^2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats why i skipped though, i was pretty sure my work til there was accurate

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now ive been doing u = pix^2 du = 2pix

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0sin(pix^2)  (x^3) = x sin(pix^2) x^2 lol.... its not that i dont trust you; i just dont trust me :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0{S} x sin(pix^2) dx ; [0,1] right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0forget the u sub for the moment....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh yeah kinda working ahead, ive done it up until here so many times haha

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh okay, well i tried by parts too, that pellet was NOT COOL

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0if this was: 2pi x sin(pix^2) you could solve it right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well yeah thats easy, because it cancels itself

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its just in terms of say udu

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.02pi [S]  x sin(pix^2) dx then multiply by a useful form of 1 lol 2pi

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0drag out the bottom 2pi and integrate it lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but i still have the 1/2pi

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no no see thats not my problem, he has the answer a 1 which is throwing me off

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0ok...let me see what I gets :) i see the issue, but not the solution :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0follow me on my u sub i have u = pix^2 du = 2pixdx dx = du/2pix

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0when i completely substitute dx from the original problem to this du/2pix, do i have to account or add anything to that? or do i just literally plug it in the equation

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0cos(pi x^2)  at 0 = 1/(2pi) 2pi cos(pi x^2)  at 1 = 1/(2pi) right? 2pi

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0nah, just plug it in; its good...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no wait dont u have to change the bounds?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.01 1  +  = pi .... so thats the answer 2pi 2pi

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0like say u OF 1 @ u = pix^2

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0only if you wanna work with u and not pix^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so after i do the integration as long as i plug u back in before i use the bounds i dont have the change them correct?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0correct; i always mess that part up so I just restsute it back :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yea i always just confuse myself and think too much into it

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0but i get pi as an answer.... as long as it was all done right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah i just got that too, he still has 1 as the answer though, pisses me off

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what goods a test review if you have the wrong answers listed!

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0tell him to prove it; show him your work for verification :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i wont see him until my exam on monday, cant really do that haha

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i know; right answers build confidence :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well i know ive done it a dozen times and come up with the same thing, so im not worried about it, hes had wrong answers before

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Just use wolfram (or equivalent) to check answer.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats the ONLY thing that was confusing me is because it didnt match my answers so i thought i did my u sub wrong

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0your u sub was fine; a little more work than I would have put into it; but fine

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i used wolfram earlier and got the same general answer with an infinite bounds

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0which is essentially the same thing

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i just read up on double ints last night; they remind me of nested functions in programing

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i wont worry about it now though however...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if you arent busy i have a long intensive problem for u lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0give me a shot at it; i gotta warm up for calc1 on monday :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0fxy if f(x,y) = sqrt(xy)  1/(((x^2)y)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ill just type it as i would say it to clarify, squareroot of both x and y, minus

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0one over x squared times y

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{\sqrt{xy}1}{x^2y}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is ONLY under the 1, nothing else

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sqrt{xy} \frac{1}{x^2y}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i was working on a dry erase board so i dont have the work anymore but i ended up doing i think 4 product rules lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0what are we to do with this function?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0x^(1/2) y^(1/2)  x^(2) y^(1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0u find der with respect to x then y

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but when you find the der with respect to y, u use what you found with respect to x

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0dF \[\frac{dF}{dy} = \frac{\sqrt{x}}{2 \sqrt{y}} + \frac{\ln(y)}{x^2}\] dx

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{dF}{dx} = \frac{\sqrt{y}}{2\sqrt{x}} + \frac{2}{xy}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0u do dx first, then when u get that der

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0prolly dint understand the questionthen lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0u do d/dy to tHAT one

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0ohh... is my dx right lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah its screwy because we thought that too, but it seemed to easy and then once u get the two different der what do u do

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hold on lemme work it again

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[1/2(xy)^{1/2}y+(x ^{2}y)^{2}2xy\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[D[(dF/dx)]/dy = \frac{1}{4\sqrt{xy}}  \frac{2x}{(xy)^2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0change everything to chain its easier to do, cause our answers arent matching up

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0working in a dim light and no paper :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the first term is good, or shoul dbe

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0but +2/x * 1/y = 2ln(y)/x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dont even worry about ln, just makes it worse than it already is, but i dont think your first is right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(xy)^1/2 in terms of X

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0Dx(sqrt(xy)) = sqrt(x)/2sqrt(y)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0x and y are transposed, but that should be it

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0Csqrt(x) = C/2sqrt(x) right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0one half goes to the front then u have (xy) then u subtract one and get (xy)^(1/2) all divided by 2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then chain the inside and get y

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the y term is constant with respect to x.... right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes thats why when u chain the inside u only get y, not x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its like 9x so to speak

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0Csqrt(x)  C/2sqrt(x)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0sqrt(y)/C > 1/C2sqrt(y) ends up 1/4sqrt(xy)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dont see where u are getting this other half from, you only get a half from the chain rule because its a sqrt

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0sqrt(xy) = sqrt(x)* sqrt(y)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, but i dont see why you would do that in the first place

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0to work them seperately; do dx first then dy and come to the proper solution..

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0that is what you said to do right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no that doesnt make any sense

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no you have to do with respect to x to the WHOLE equation first

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0once you find with respect to x u hvae to apply with respect to y after that

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0oh, like implicit then?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0then what you want is to hold y constant and dx first right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes i thought we already said that earlier, y is just y, it does nothing now

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0so i split the sqrt to work with only the x

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0sqrt(y)  2sqrt(x)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0you say no...but tell me why not?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay we have the orignal problem blah blah

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we take the derivative of that with ONLY respect to x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0once we get that function we take the derivative of that with respect to y

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0and since it is a sum we can do each term seperately...lets work the first term.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0sqrt(xy) = sqrt(x) * sqrt(y) by definition that is appropriate

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0u just chain rule so .5(xy)^(1/2) times y

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0just look at sqrt(xy) as (xy)^(1/2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now instead of splitting stuff up, u just made the whole thing a chain rule which is reaaaaaaaaally easy

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you kno what chain rule is right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0sqrt(Cx) >> C/2sqrt(Cx) right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because its just (Cx)^(1/2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do you kno what the chain rule is though?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0so that translates to: y  correct? 2sqrt(xy)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0lol...no, i just do double integrals for a hobby ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well with this problem its easier if you just put everything to a negative exponent

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its verrrrrrrry long, and that makes it faster, especially when u are trying to do it in terms of x or y

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0easier is a subjective term;

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well bringing something to the front, minus 1, then der of the inside is pretty easy, dont think thats subjective

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0so we got: y  and should dy that now right? 2sqrt(xy)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no u still have to do dx to the last part, that was only the first part lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the last term can wait; its part of another section of the addition

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the eq is \[(xy)^{1/2}1/(x ^{2}y)\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0right; so the first term is derivable n its own right without the second

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you can, you are going to have to end up doing the second, and for thought it seems simpler to just do all the dx first

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now you just product the 1st term get y, then product the second term

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the length of this posting is messing up my system :) ill let you have at it alone ;)
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