I don't know where to start
#Reverse order of integration
normal fox
jade reef
alright so first off you want to graph your limits
starting with 8y=x as your lower and 8=X for your upper, then you want to graph 0=y for your lower and 1=y for your upper, which looks like this: (now at the bottom of my message)
where you know your lower is 8y and your upper is 8, so you will be integrating in that bottom triangle
now you flip your order to get ∫∫e^x^2dydx and must find your limits, going in the y direction your first limit is 0, and then next thing you'll hit is the line ⅛x=y so you have your y limits 0→⅛x
now for your X limits, the smallest value is X=0 and the largest is X=8, so you now have your limits for X and y
∫0→8∫0→⅛x(e^x^2)dydx
now you have ∫0→8(ye^x^2|0→⅛x)dx
so ∫0→8(⅛xe^x^2)dx
=⅛∫0→8(xe^x^2)dx
now
u=x²
du=2xdx
(1/16)∫0→64(e^(u))du (change of limits)
(1/16)[e^u|0→64]=(1/16)(e⁶⁴-e⁰)=(1/16)(e⁶⁴-1)
tadah
I'd you have any questions just ask and I'll answer to the best of my ability
@normal fox
tbh I'm kinda peeved that they've only given 1 mark to that question, but I can understand that it isn't that complicated once you understand the motions of reversing the order
normal fox
tbh I'm kinda peeved that they've only given 1 mark to that question, but I can ...
each hw was worth 1 point so its understandable. If it was a test and it was worth 1 point that would be outrageous
jade reef
each hw was worth 1 point so its understandable. If it was a test and it was wor...
ehh I could understand 2 points but 1 point would be pushing it
depends which year of uni you're in
mark inflation
