Calculate the double integral.

Question

please help me

Answer 1

∫∫xy²/(1 + x²)dxdy = ∫₀¹x/(1 + x²)dx∫_-3³y²dy = 1/2ln(1 + x²)₀¹·2y³/3₀³ = (ln2 - ln1)(9 - 0) = 9ln2

Answer 2

∫_{(from y = -3 to 3)} [∫_{(from x = 0 to 1)}((xy²) / (x²+1))dx]dy

= ∫_{(from y = -3 to 3)} y² [∫_{(from x = 0 to1)}(x / (x²+1))dx] dy (treat y as constant in the inner integral)

= ∫_{(from -3 to 3)} y²[(1/2)ln(x²+1)]_{(from x = 0 to 1)}dy

= (1/2)∫_{(from -3 to 3)}y²(ln2 - ln1)dy

= ( (ln2) / 2)(1/3)y³_{(from -3 to 3)} = (1/6)ln2(27 - (-27)) = (54/6)ln2 = 9ln2 = ln512

Answer 3

What makes this problem fairly easy is that there is no mixture of x and y terms or a non-rectangular region.

Using I[a,b] for the integral from a to b and E[a,b] for the evaluation from a to b, this is simply

I[0,1]I[-3,3]xy^2/(x^2+1) dy dx =

I[0,1] x/(x^2+1) y^3/3 E[-3,3] dx =

1/3 I[0,1] x/(x^2+1)( (3^3) - ((-3)^3)) dx =

18 I[0,1] x/(x^2+1) dx =

9 ln(x^2 + 1) E[0,1] =

9 ln(2) - 9 ln(1) =

9 ln(2)

Answer 4

= ∫ x from 0 to 1 ∫ y from -3 to 3 of xy² dx dy / (x² + 1)

= ∫ x from 0 to 1 [y³ / 3, with y from -3 to 3] x dx / (x² + 1)

= ∫ x from 0 to 1 [3³ / 3 - (-3)³ / 3] x dx / (x² + 1)

= ∫ x from 0 to 1 [27/3 - (-27)/3] x dx / (x² + 1)

= ∫ x from 0 to 1 [9 - (-9)] x dx / (x² + 1) = ∫ x from 0 to 1 of 18x dx / (x² + 1),

but now if we let u = x² + 1, then du = 2x dx, so 18x dx = 9 du;

when x = 0, u = 0² +1 = 0 + 1 = 1; and when x = 1, u = 1² + 1 = 1 + 1 = 2,

so now we have the following integral to calculate:

∫ u from 1 to 2 of 9 du / u = 9 ln(u), with u from 1 to 2 = 9 ln(2) - 9 ln(1)

= 9 ln(2) - 9*0 = 9 ln(2) - 0 = 9 ln(2).