## anonymous 3 years ago double integral question.

1. anonymous

2. amistre64

its nice they give you a box to work in

3. amistre64

let y+1 = some constant, say k $\int_{0}^{2}\frac{1}{(x+k)^3}dx$ $\int_{0}^{2}(x+k)^{-3}dx$

4. amistre64

whats your solution to this part?

5. anonymous

|dw:1370890838800:dw|

6. experimentX
7. anonymous

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

8. anonymous

i forgot the (-) on the 2 up there

9. amistre64

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. amistre64

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

11. amistre64

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

12. anonymous

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

13. amistre64

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

14. anonymous

why not with respect to "y"?

15. amistre64

because your order of integration is dxdy dx first, then dy

16. amistre64

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

17. amistre64

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

18. anonymous

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

19. amistre64

you most likely added things incorrectly

20. amistre64

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

21. anonymous

ouch!...i am redoing this thing!

22. amistre64

:) the algebra/arithmetic is usually the culprit

23. anonymous

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

24. anonymous

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

25. amistre64

:) good job