Question

if you have integral of 1/(2x+3y)^5 dA R=[0.1]*[1,2] how do we solve it

Answer 1

R=[0.1]*[1,2] ??

Answer 2

this must be a definite integral ... what are the limits for x and y??

Answer 3

iterated integral
limits for x are [0,1] and y are [1,2]

Answer 4

change to polar coordinates

Answer 5

how

Answer 6

\( x = r \cos \theta \; \; and \; \; y = r \sin \theta \)

Answer 7

and get rid to those 2's and 3's

Answer 8

and plot those points .... on your graph paper .. get limits for r and \theta

Answer 9

\[\int\limits_{0}^{1} \int\limits_{1}^{2} 1/(2x+3y)^2 dydx\] this is the equation but what happens when the power is changed

Answer 10

Jeez ... you were making things complicated on your own???
I don't know ... for 2 i guess you will get the answer in same pattern ... physically ... they are whole lotta different ... can't say anything about the answer without doing it first

Answer 11

how do i go abt doing it

Answer 12

put x and y as i mentioned above .... and dA = r dr d\theta

though i doubt if you can do in using rectangular coordinates

Answer 13

i took the (2x+3y)^2 up to give me (2x+3y)^-2 whats the next step

Answer 14

not sure about rectangular coordinates ...
this is how i do it in polar coordinates

\( 2x +3y = 2r\cos \theta + 3 r \sin\theta = r(2\cos\theta + 3\sin\theta) = r A\cos(\theta - \phi)\)
where, \( A = \sqrt{2^2 + 3^3} \) and \( \phi = \tan^{-1} (2/3) \)

Answer 15

im not allowed to use polar coordinates we have not reached that stage apparently its rectangular coordinates

Answer 16

I thought you would do polar coordinate on High school and double integrals on undergraduate ...
well ... let's see

Answer 17

first integrate x or y treating other as constant!!

Answer 18

final answer should be [-1/6 (\ln 8-\ln 2-\ln 5)\]

Answer 19

Looks like pretty simple ...
http://www.wolframalpha.com/input/?i=integrate+%282x+%2B+3b%29^-2+dx
I was the one making things unnecessarily complicated

Answer 20

will check it out its still loading

Answer 21

evaluate this first
\[ \int_0^1 \frac{1}{(3x+2y)^2} dx\]

Answer 22

let \( 2x+3y = u, du = 3dx\) then we have \( u = 3y \) to \( 3y + 4\)
\[ \frac{1}{3}\int_{3y}^{3y+4} \frac{du}{u^2}\]

Answer 23

Oops ... sorry about the that ... 1/3 was supposed to be 1/2 ... I made mistake in copying the integral ... in the post two steps above

Answer 24

im lost let me redo it ur way

Answer 25

its 2x+3y

Answer 26

k thnx

Answer 27

y is ur du=3dx

Answer 28

because i supposed 2x+3y=u, 2dx=du <--- it was supposed to be ... i made error copying!!

Answer 29

ok where is the second u coming from?

Answer 30

?? can you use draw function to demonstrate what you mean or ... clarify what you are trying to say??

Answer 31

let 2x+3y=u,du=3dx then we have u=3y to 3y+4
that u=3y where do u get it bcz when i substitute back i get \[1/2 \int\limits_{0}^{1} 1/u^2 du\]

Answer 32

what do i do after that step whats the integral of 1/u^2

Answer 33

\[ \left [ \frac{u^{-1}}{-1}\right ]_{3y}^{3y+4} = \frac{1}{3y} - \frac{1}{3y+4}\]
Integrate it again with respect to y
\[ \int_1^2 \left ( \frac{1}{3y} - \frac{1}{3y+4} \right )dy\]

Answer 34

Integrate 1/u^2 within those limits ... and get the result in y, and integrate it with dy again!!
the latter one is pretty straightforward!!

Answer 35

i get [1/2[-1/(2+3y) +1/3y]

Answer 36

Oh ... yes ... i put the wrong limit, for x .. i guess that is right .. you can do the rest.

Answer 37

im stuck

Answer 38

because when i continue i dont get the same eanswer

Answer 39

try wolfram

Answer 40

k let me simplify if then i will tell you what i get

Answer 41

http://www.wolframalpha.com/input/?i=integrate+%282x+%2B+3y%29^-2+dx+from+x%3D0+to+1

Answer 42

http://www.wolframalpha.com/input/?i=integrate+1%2F%286+y%2B9+y^2%29dy+from+1+to+2

Answer 43

wolfram seems to comply with your answer
unless you or I made mistake in integration ... you should the answer!

Answer 44

i give up #sad