Evaluate the sum of these two integrals [0 to 6] ∫[0 to x/3] (x^2 + y^2) dy dx + ∫[0 to 6] ∫[x/3 to 2] (x^2 + y^2) dy dx

Answer 1

Let's evaluate the integral step by step:

First, let's evaluate the integral [0 to x/3] ∫(x^2 + y^2) dy.

Integrating (x^2 + y^2) with respect to y, we get:
∫(x^2 + y^2) dy = xy + (1/3)y^3

Now, evaluate this expression at the upper limit x/3 and lower limit 0:
= x * (x/3) + (1/3) * (x/3)^3
= x^2/3 + (1/3) * (x^3/27)
= x^2/3 + x^3/81

Now, evaluate this expression at the upper limit 6 and lower limit 0:
= 6^2/3 + 6^3/81 - 0^2/3 - 0^3/81
= 12 + 432/81
= 12 + 16/3

= 12 + 5.3333
= 17.3333

Now, let's evaluate the second integral, ∫[x/3 to 2] (x^2 + y^2) dy.

Integrating (x^2 + y^2) with respect to y, we get:
∫(x^2 + y^2) dy = xy + (1/3)y^3

Now, evaluate this expression at the upper limit 2 and lower limit x/3:
= 2x + (1/3)(2^3 - (x/3)^3)
= 2x + (8/3 - x^3/27)
= 2x + 24/3 - x^3/27
= 2x + 8 - x^3/27

Now, let's evaluate this integral from x = 0 to x = 6:

∫[0 to 6] (2x + 8 - x^3/27) dx

Integrating 2x with respect to x, we get:
= x^2

Integrating 8 with respect to x, we get:
= 8x

Integrating -x^3/27 with respect to x, we get:
= -(1/27)(1/4)x^4

Now, evaluate these expressions at the upper limit 6 and lower limit 0:

= (6^2) + 8(6) - (1/27)(1/4)(6^4) - (0^2) - 8(0) + (1/27)(1/4)(0^4)
= 36 + 48 - (1/27)(1/4)(1296) - 0 - 0 + 0
= 84 - (1296/108) - 0
= 84 - 12
= 72

Thus, the sum of the two integrals is 17.3333 + 72 = 89.3333.