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tomikoBest ResponseYou've already chosen the best response.0
can someone please help me solve this? http://gyazo.com/d88e00524337c49ec6734dcb2b9ad5b1
 10 months ago

amistre64Best ResponseYou've already chosen the best response.2
its nice they give you a box to work in
 10 months ago

amistre64Best ResponseYou've already chosen the best response.2
let y+1 = some constant, say k \[\int_{0}^{2}\frac{1}{(x+k)^3}dx\] \[\int_{0}^{2}(x+k)^{3}dx\]
 10 months ago

amistre64Best ResponseYou've already chosen the best response.2
whats your solution to this part?
 10 months ago

Euler271Best ResponseYou've already chosen the best response.0
dw:1370890838800:dw
 10 months ago

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

tomikoBest ResponseYou've already chosen the best response.0
i forgot the () on the 2 up there
 10 months ago

amistre64Best 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 \[\int_{a}^{b}f(x)~dx=F(b)F(a)\]
 10 months ago

amistre64Best 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
 10 months ago

amistre64Best ResponseYou've already chosen the best response.2
\[\frac12\left(\int_{0}^{1}(y+1)^{2}dy\int_{0}^{1}(y+3)^{2}dy\right)\]
 10 months ago

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

amistre64Best 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
 10 months ago

tomikoBest ResponseYou've already chosen the best response.0
why not with respect to "y"?
 10 months ago

amistre64Best ResponseYou've already chosen the best response.2
because your order of integration is dxdy dx first, then dy
 10 months ago

amistre64Best 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 \[...\left(\int_{0}^{2}\frac{1}{(x+k)^3}~dx\right)...\]
 10 months ago

amistre64Best 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 \[...\left(\frac{1}{2(2+y+1)^2}+\frac{1}{2(y+1)^2}\right)...\] \[\frac12\int_{0}^{1}\left(\frac{1}{(y+1)^2}\frac{1}{(y+3)^2}\right)dy\]
 10 months ago

tomikoBest ResponseYou've already chosen the best response.0
hmmm....i got 1/24 as final answer....
 10 months ago

amistre64Best ResponseYou've already chosen the best response.2
you most likely added things incorrectly
 10 months ago

amistre64Best ResponseYou've already chosen the best response.2
\[\frac12\int_{0}^{1}\left(\frac{1}{(y+1)^2}\frac{1}{(y+3)^2}\right)dy\] \[\frac12\left(\frac{1}{(y+1)}\frac{1}{(y+3)}\right)\] \[\frac12\left(\frac{1}{(1+1)}\frac{1}{(1+3)}\right)\frac12\left(\frac{1}{(0+1)}\frac{1}{(0+3)}\right)\] \[\frac12\left(\frac14\frac12\right)\frac12\left(\frac131\right)\] \[\frac12\left(\frac14\frac12\frac13+1\right)\] \[\frac12\left(\frac{364+12}{12}\right)\] \[\frac12\left(\frac{5}{12}\right)\]
 10 months ago

tomikoBest ResponseYou've already chosen the best response.0
ouch!...i am redoing this thing!
 10 months ago

amistre64Best ResponseYou've already chosen the best response.2
:) the algebra/arithmetic is usually the culprit
 10 months ago

tomikoBest ResponseYou've already chosen the best response.0
exactly. i did it a second time and got 11/24. i'm doing it again!!
 10 months ago

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