Double Integration?

Question

So how do I work this?

(1 to 2)(y to y^3) e^(x/y) dx dy

the 1 to 2 is the vetical squigly line for integral.

any help would be appreciated, thanks

the y to y^3 is also one of the integral symbols too.

ok so i got everything up till

∫ye^(y²)=(1/2)e^(y^2)

what happened to the y infront? this is where i got stuck on the problem, im good up till then and after. i think im forgetting a rule.

OOHHHH!!!! i see it now, its soo simple, i cant beleive i missed that!

Answer 1

Do the inner integral first:

∫e^(x/y)dx
Treat y as a constant
∫e^(x/y)dx
= ye^(x/y)

Apply the limits:
ye^(y³/y) - ye^(y/y)
= ye^(y²) - ye^(1)

Integrate this wrt y:
∫ye^(y²) - ye^(1)dy
= (1/2)e^(y²) - (y²/2)e^(1)

Apply the limits:
(1/2)e^(2²) - (2²/2)e^(1) - [(1/2)e^(1²) - (1²/2)e^(1)]
= (1/2)e^(4) - 2e^(1) - (1/2)e^(1) + (1/2)e^(1)
= (1/2)e^(4) - 2e^(1)

Edit:
∫ye^(y²)dy
y² = u
2ydy = du
∫ye^(y²)dy
= ∫(1/2)(e^u)du
= (e^u)/2
= (1/2)e^(y²)

Answer 2

=Int (1,2)dy * Int (y,y^3) e^(x/y) dx
Int(y,y^3) e^(x/y) dx =y(e^x/y) between x= y^3 and x=y
= y( e^y^2-e)
now we have Int (1,2)(-e y+ye^y^2)dy = -e/2*y^2 +1/2e^y^2 (1,2) =-2e +1/2 e^4 +e/2- 1/2e = -2e+1/2e^4

Answer 3

♠ = 21.86; are you deleting your questions?