Question

To evaluate the double integral ∬R1+x2yx3dA over the rectangle R=[0,1]×[0,1], you can perform the following calculations:

∬R1+x2yx3dA=∫01∫011+x2yx3dydx

Answer 1

To evaluate the double integral ∬R(1+x^2)(yx^3)dA over the rectangle R=[0,1]×[0,1], you can perform the following calculations: ∬R(1+x^2)(yx^3)dA=∫₀¹∫₀¹(1+x^2)(yx^3)dydx.

First, integrate the inner integral with respect to y from 0 to 1: ∫₀¹(1+x^2)(yx^3)dy = [(1+x^2)/2](x^3/4) = x^3(1+x^2)/8.

Then, integrate the outer integral with respect to x from 0 to 1: ∫₀¹x^3(1+x^2)/8 dx = [x^4/32 + x^6/24] from 0 to 1.

Substituting the limits of integration: [(1/32 + 1/24) - (0/32 + 0/24)] = 1/32 + 1/24 = 5/96.

Step-by-step explanation:

To evaluate the double integral ∬R(1+x^2)(yx^3)dA over the rectangle R=[0,1]×[0,1], you can perform the following calculations: ∬R(1+x^2)(yx^3)dA=∫₀¹∫₀¹(1+x^2)(yx^3)dydx.

First, integrate the inner integral with respect to y from 0 to 1:

∫₀¹(1+x^2)(yx^3)dy = [(1+x^2)/2](x^3/4) = x^3(1+x^2)/8.

Then, integrate the outer integral with respect to x from 0 to 1:

∫₀¹x^3(1+x^2)/8 dx = [x^4/32 + x^6/24] from 0 to 1.

Substituting the limits of integration:

[(1/32 + 1/24) - (0/32 + 0/24)] = 1/32 + 1/24 = 5/96.