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

Question

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.

Answer 2

To evaluate the sum of the two integrals, let's start with the first integral:

∫[0 to x/3] (x^2 + y^2) dy

We can integrate this with respect to y while treating x as a constant:

∫[0 to x/3] (x^2 + y^2) dy = x^2y + (y^3 / 3) |[0 to x/3]
= (x^3 / 3) + (x^3 / 27)
= (10x^3 / 27) (simplified)

Next, let's move on to the second integral:

∫[x/3 to 2] (x^2 + y^2) dy dx

Here, we integrate with respect to y first, treating x as a constant:

∫[x/3 to 2] (x^2 + y^2) dy = x^2y + (y^3 / 3) |[x/3 to 2]
= 2x^2 + (8/3 - x^3/27)

Now, we integrate this expression with respect to x:

∫[0 to 6] ∫[x/3 to 2] (x^2 + y^2) dy dx = ∫[0 to 6] (2x^2 + (8/3 - x^3/27)) dx
= (2/3)x^3 + (8/3)x - (1/108)x^4 |[0 to 6]
= 144 + 16 - 2/3 - 0
= 952/3 (simplified)

Finally, to find the sum of the two integrals, we add them together:

Sum = (10x^3 / 27) + (952/3)
= (10x^3 / 27) + (952/3)

So the sum of the two integrals is (10x^3 / 27) + (952/3).