how to evaluate the following integral: ∫ (0 to 1)∫ sqrt(y) to 1 e.^x.^3 dx dy?
回答 (2)
2014-08-26 5:07 pm
The region of integration as given is x = y^(1/2) to x = 1 with y in [0, 1].
After sketching this, observe that this can be rewritten as
y = 0 to y = x^2 with x in [0, 1].
So, the integral equals
∫(x = 0 to 1) ∫(y = 0 to x^2) e^(x^3) dy dx
= ∫(x = 0 to 1) x^2 e^(x^3) dx
= (1/3)e^(x^3) {for x = 0 to 1}
= (1/3)(e - 1).
I hope this helps!
After sketching this, observe that this can be rewritten as
y = 0 to y = x^2 with x in [0, 1].
So, the integral equals
∫(x = 0 to 1) ∫(y = 0 to x^2) e^(x^3) dy dx
= ∫(x = 0 to 1) x^2 e^(x^3) dx
= (1/3)e^(x^3) {for x = 0 to 1}
= (1/3)(e - 1).
I hope this helps!
2014-08-26 1:00 pm
Integrate e^(3x)
e^(3x)/3
Bounds
e^(3sqrt(y))-(e^3)/3
u=sqrt(y)
Du = 1/(2sqrt(y)) dy
Du=1/(2u) dy
Dy= 2u du
e^(3u)(2u) du - (e^3)/3 dy
Integrate by parts
Let a=2u, db=e^(3u)
Da=2, b=(e^3u)/3
(2/3)u*e^3 - integral((2/3)e^3u) - ((e^3)/3)y
(2/3)sqrt(y)e^3 - (2/9)e^(3sqrt(y)) - ((e^3)/3)y
Bounds
(2/3)e^3 - (2/9)e^3 - (1/3)e^3 -(2/9)
(1/9)e^3 - 2/9
e^(3x)/3
Bounds
e^(3sqrt(y))-(e^3)/3
u=sqrt(y)
Du = 1/(2sqrt(y)) dy
Du=1/(2u) dy
Dy= 2u du
e^(3u)(2u) du - (e^3)/3 dy
Integrate by parts
Let a=2u, db=e^(3u)
Da=2, b=(e^3u)/3
(2/3)u*e^3 - integral((2/3)e^3u) - ((e^3)/3)y
(2/3)sqrt(y)e^3 - (2/9)e^(3sqrt(y)) - ((e^3)/3)y
Bounds
(2/3)e^3 - (2/9)e^3 - (1/3)e^3 -(2/9)
(1/9)e^3 - 2/9
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