Question

double integral question.

Answer 1

can someone please help me solve this? http://gyazo.com/d88e00524337c49ec6734dcb2b9ad5b1

Answer 2

its nice they give you a box to work in

Answer 3

let y+1 = some constant, say k

\[\int_{0}^{2}\frac{1}{(x+k)^3}dx\]
\[\int_{0}^{2}(x+k)^{-3}dx\]

Answer 4

whats your solution to this part?

Answer 5

http://www.wolframalpha.com/input/?i=Integrate%5BIntegrate%5B1%2F%28x%2By%2B1%29%5E3%2C+%7Bx%2C+0%2C+2%7D%5D%2C%7By%2C+0%2C+1%7D%5D

Answer 6

@amistre64 solution \[-\frac{ (x + k) ^ 2 }{ 2} + C\]

Answer 7

i forgot the (-) on the 2 up there

Answer 8

that would be the indefinite integral yes; lets apply the limits of 0 to 2, recall that \[\int_{a}^{b}f(x)~dx=F(b)-F(a)\]

Answer 9

\[\left(-\frac{ (2 + k) ^{-2} }{ 2}\right)-\left(-\frac{ (0 + k) ^{-2} }{ 2} \right)\]
\[-\frac{ (2 + k) ^{-2} }{ 2}+\frac{ ( k) ^{-2} }{ 2} \]
\[\frac12(k^{-2}-(2 + k) ^{-2})~:~k=y+1 \]
\[\frac12((y+1)^{-2}-(y+3) ^{-2})\]
integrate that from 0 to 1

Answer 10

\[\frac12\left(\int_{0}^{1}(y+1)^{-2}dy-\int_{0}^{1}(y+3)^{-2}dy\right)\]

Answer 11

from your 2nd step, am i not to replace k = 2? and k = 0?
@amistre64

Answer 12

no, k is just some constant with respect to x your integrating "x" from 0 to 2 .... not "k" from 0 to 2

Answer 13

why not with respect to "y"?

Answer 14

because your order of integration is dxdy

dx first, then dy

Answer 15

\[\int_{0}^{1}\left(\int_{0}^{2}\frac{1}{(x+y+1)^3}~dx\right)dy\]
the inside stuff is dx .. anything that is not xed is constant, therefore y+1 = k
\[...\left(\int_{0}^{2}\frac{1}{(x+k)^3}~dx\right)...\]

Answer 16

\[...\left(\int_{0}^{2}\frac{1}{(x+k)^3}~dx\right)...\]
\[...\left(-\frac{1}{2(2+k)^2}+\frac{1}{2(0+k)^2}\right)...\]
replace k with y+1
\[...\left(-\frac{1}{2(2+y+1)^2}+\frac{1}{2(y+1)^2}\right)...\]
\[\frac12\int_{0}^{1}\left(\frac{1}{(y+1)^2}-\frac{1}{(y+3)^2}\right)dy\]

Answer 17

hmmm....i got 1/24 as final answer....

Answer 18

you most likely added things incorrectly

Answer 19

\[\frac12\int_{0}^{1}\left(\frac{1}{(y+1)^2}-\frac{1}{(y+3)^2}\right)dy\]
\[\frac12\left(\frac{1}{-(y+1)}-\frac{1}{-(y+3)}\right)\]
\[\frac12\left(\frac{1}{-(1+1)}-\frac{1}{-(1+3)}\right)-\frac12\left(\frac{1}{-(0+1)}-\frac{1}{-(0+3)}\right)\]
\[\frac12\left(\frac14-\frac12\right)-\frac12\left(\frac13-1\right)\]
\[\frac12\left(\frac14-\frac12-\frac13+1\right)\]
\[\frac12\left(\frac{3-6-4+12}{12}\right)\]
\[\frac12\left(\frac{5}{12}\right)\]

Answer 20

ouch!...i am redoing this thing!

Answer 21

:) the algebra/arithmetic is usually the culprit

Answer 22

exactly. i did it a second time and got 11/24. i'm doing it again!!

Answer 23

@amistre64 got it finally. my friend also got 5/24 :). thank you very much!!

Answer 24

:) good job