We can start by evaluating the inner integral first with respect to y while treating x as a constant:
∫(2,1) e^(y/x) dy evaluated at y=x^3 and y=x
= ∫(2,1) e^(y/x) dx^3 - ∫(2,1) e^(y/x) dx
= x^3(e - e^(8/x))/x - (e - e^(2/x))
= x^2(e - e^(8/x)) - (e - e^(2/x))
Next, we can integrate this expression with respect to x:
∫(1,2) x^2(e - e^(8/x)) - (e - e^(2/x)) dx
= [1/3 x^3 (e - e^(8/x)) - x(e - e^(2/x))] evaluated at x=1 and x=2
= [1/3 (2^3 - 1^3)(e - e^(8/2)) - 2(e - e^(2/2))] - [1/3 (1^3 - 1^3) (e - e^(8/1)) - 1(e - e^(2/1))]
= 2/3(e - e^4) - 2e + 2 - 1/3(e - e^8) + 1(e - 1)
= 2/3 - 1/3 e^8 + 8/3 e^4 - 2e + 1
= -2/3 + 8/3 e^4 - 1/3 e^8
Therefore, the final answer is -2/3 + 8/3 e^4 - 1/3 e^8.
Evaluate iterated integral, ∫(2,1) ∫(x^3,x) e^(y/x) dydx
3 answers
AAAaannndd the bot gets it wrong yet again!
∫[1..2] ∫[x..x^3] e^(y/x) dy dx
= ∫[1..2] x(e^(x^2)-e) dx
= 1/2 (e^x^2 - ex^2) [1..2]
= e/2 (e^3 - 4)
∫[1..2] ∫[x..x^3] e^(y/x) dy dx
= ∫[1..2] x(e^(x^2)-e) dx
= 1/2 (e^x^2 - ex^2) [1..2]
= e/2 (e^3 - 4)
I apologize for the error in my previous response. Thank you for providing the correct solution. You are correct that the iterated integral evaluates to:
∫[1..2] ∫[x..x^3] e^(y/x) dy dx
= ∫[1..2] x(e^(x^2)-e) dx
= 1/2 (e^x^2 - ex^2) [1..2]
= e/2 (e^3 - 4)
∫[1..2] ∫[x..x^3] e^(y/x) dy dx
= ∫[1..2] x(e^(x^2)-e) dx
= 1/2 (e^x^2 - ex^2) [1..2]
= e/2 (e^3 - 4)