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tomiko
 2 years ago
double integral question.
tomiko
 2 years ago
double integral question.

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tomiko
 2 years ago
Best ResponseYou've already chosen the best response.0can someone please help me solve this? http://gyazo.com/d88e00524337c49ec6734dcb2b9ad5b1

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.2its nice they give you a box to work in

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.2let y+1 = some constant, say k \[\int_{0}^{2}\frac{1}{(x+k)^3}dx\] \[\int_{0}^{2}(x+k)^{3}dx\]

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.2whats your solution to this part?

tomiko
 2 years ago
Best ResponseYou've already chosen the best response.0@amistre64 solution \[\frac{ (x + k) ^ 2 }{ 2} + C\]

tomiko
 2 years ago
Best ResponseYou've already chosen the best response.0i forgot the () on the 2 up there

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.2that would be the indefinite integral yes; lets apply the limits of 0 to 2, recall that \[\int_{a}^{b}f(x)~dx=F(b)F(a)\]

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.2\[\left(\frac{ (2 + k) ^{2} }{ 2}\right)\left(\frac{ (0 + k) ^{2} }{ 2} \right)\] \[\frac{ (2 + k) ^{2} }{ 2}+\frac{ ( k) ^{2} }{ 2} \] \[\frac12(k^{2}(2 + k) ^{2})~:~k=y+1 \] \[\frac12((y+1)^{2}(y+3) ^{2})\] integrate that from 0 to 1

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.2\[\frac12\left(\int_{0}^{1}(y+1)^{2}dy\int_{0}^{1}(y+3)^{2}dy\right)\]

tomiko
 2 years ago
Best ResponseYou've already chosen the best response.0from your 2nd step, am i not to replace k = 2? and k = 0? @amistre64

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.2no, k is just some constant with respect to x your integrating "x" from 0 to 2 .... not "k" from 0 to 2

tomiko
 2 years ago
Best ResponseYou've already chosen the best response.0why not with respect to "y"?

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.2because your order of integration is dxdy dx first, then dy

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.2\[\int_{0}^{1}\left(\int_{0}^{2}\frac{1}{(x+y+1)^3}~dx\right)dy\] the inside stuff is dx .. anything that is not xed is constant, therefore y+1 = k \[...\left(\int_{0}^{2}\frac{1}{(x+k)^3}~dx\right)...\]

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.2\[...\left(\int_{0}^{2}\frac{1}{(x+k)^3}~dx\right)...\] \[...\left(\frac{1}{2(2+k)^2}+\frac{1}{2(0+k)^2}\right)...\] replace k with y+1 \[...\left(\frac{1}{2(2+y+1)^2}+\frac{1}{2(y+1)^2}\right)...\] \[\frac12\int_{0}^{1}\left(\frac{1}{(y+1)^2}\frac{1}{(y+3)^2}\right)dy\]

tomiko
 2 years ago
Best ResponseYou've already chosen the best response.0hmmm....i got 1/24 as final answer....

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.2you most likely added things incorrectly

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.2\[\frac12\int_{0}^{1}\left(\frac{1}{(y+1)^2}\frac{1}{(y+3)^2}\right)dy\] \[\frac12\left(\frac{1}{(y+1)}\frac{1}{(y+3)}\right)\] \[\frac12\left(\frac{1}{(1+1)}\frac{1}{(1+3)}\right)\frac12\left(\frac{1}{(0+1)}\frac{1}{(0+3)}\right)\] \[\frac12\left(\frac14\frac12\right)\frac12\left(\frac131\right)\] \[\frac12\left(\frac14\frac12\frac13+1\right)\] \[\frac12\left(\frac{364+12}{12}\right)\] \[\frac12\left(\frac{5}{12}\right)\]

tomiko
 2 years ago
Best ResponseYou've already chosen the best response.0ouch!...i am redoing this thing!

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.2:) the algebra/arithmetic is usually the culprit

tomiko
 2 years ago
Best ResponseYou've already chosen the best response.0exactly. i did it a second time and got 11/24. i'm doing it again!!

tomiko
 2 years ago
Best ResponseYou've already chosen the best response.0@amistre64 got it finally. my friend also got 5/24 :). thank you very much!!
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