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

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

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

Euler271
 one year ago
Best ResponseYou've already chosen the best response.0dw:1370890838800:dw

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

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

amistre64
 one year 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
 one year 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
 one year 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
 one year 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
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.0why not with respect to "y"?

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

amistre64
 one year 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
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.0hmmm....i got 1/24 as final answer....

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

amistre64
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.0ouch!...i am redoing this thing!

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

tomiko
 one year 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
 one year 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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