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

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

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

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

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

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

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

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

amistre64 Group TitleBest 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
 one year ago

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

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

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

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

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

amistre64 Group TitleBest 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 \[...\left(\int_{0}^{2}\frac{1}{(x+k)^3}~dx\right)...\]
 one year ago

amistre64 Group TitleBest 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 \[...\left(\frac{1}{2(2+y+1)^2}+\frac{1}{2(y+1)^2}\right)...\] \[\frac12\int_{0}^{1}\left(\frac{1}{(y+1)^2}\frac{1}{(y+3)^2}\right)dy\]
 one year ago

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

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

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

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

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

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

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

amistre64 Group TitleBest ResponseYou've already chosen the best response.2
:) good job
 one year ago
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