Evaluate the integral by reversing the order of integration.
(2 integrations)
0 to 1
sqrt(x) to 1
8(y^3+1) dy dx
∫[0,1]∫[√x,1] 8(y^3+1) dy dx
= ∫[0,1]8(y^4/4+y)|[√x,1] dx
= ∫[0,1]8[(1/4+1)-(x^2/4+√x)] dx
= ∫[0,1] -2x^2 - 8√x + 10 dx
= (-2/3 x^3 - 16/3x^(3/2) + 10x [0,1]
= -2/3 - 16/3 + 10
= 4
To evaluate the integral by reversing the order of integration, we need to swap the order of the variables, which means converting the given integral from an iterated integral with respect to y first, then x, into an iterated integral with respect to x first, then y.
The original integral is:
∫∫[0 to 1] [√(x) to 1] 8(y^3 + 1) dy dx
To reverse the order of integration, we need to visualize the region of integration. In this case, the region is bounded by the curves y = √(x) and y = 1, and the lines x = 0 and x = 1.
By drawing a diagram or referring to the graphs of the curves, we determine that the region is a triangular shape between the y-axis, the line y = 1, and the curve y = √(x).
Now, let's reverse the order of integration:
∫∫[0 to 1] [√(x) to 1] 8(y^3 + 1) dy dx
∫∫[0 to 1] [0 to y^2] 8(y^3 + 1) dx dy
Now, we integrate with respect to x first:
∫[0 to y^2] 8(y^3 + 1) dx
= 8(y^3 + 1) x ∣ [0 to y^2]
= 8(y^5 + y^2)
Now, we integrate with respect to y:
∫[0 to 1] 8(y^5 + y^2) dy
= 8/6 (y^6/6 + y^3/3) ∣ [0 to 1]
= 8/6 (1/6 + 1/3)
= 8/6 (1/6 + 2/6)
= 8/6 (3/6)
= 1/3
Therefore, the value of the integral by reversing the order of integration is 1/3.