can u help me on this prob?

Question

evaluate the iterated integral
....ln6........5
I= integ (( integ e^(2x-y) ) dy
....0...........0

what would I=?

Answer 1

You are missing dx. Since dy is outside the parentheses, I assume dx was meant to be inside the first integral. To integrate with respect to one variable when other variables are present, just treat the other variables as constants.

Integrating (e^(2x - y) with respect to dx yields

(1/2)e^(2x - y) + C,

which evaluated from x = 0 to 5 is

(1/2)e^(10 - y) + (1/2)e^(-y).

Integrating this with respect to dy yields

(-1/2)e^(10 - y) - (1/2)e^(-y) + C,

which evaluated from 0 to ln6 is

(-1/2)e^(10 - ln6) + (1/2)e^10 - (1/2)e^(-ln6) + (1/2)e^0 =
(-1/2)((e^10) / 6) + (1/2)e^10 - (1/2)(1/6) + (1/2)(1) =
(-1/12)e^10 + (1/2)e^10 - (1/12) =
(5/12)e^10 - (1/12) =
((e^10) - 1) / 12.