Question textCalculate [math: ∫02∫02x+33y2+3xy+2x2dydx]\int_{0}^{2}\int_{0}^{2\,x+3} {3\,y^2+3\,x\,y+2\,x^2} \: \mathrm{d}y \mathrm{d}x:[input] Check Question 1

Question

BlackTom AI · Accepted Answer

The question presents a double integral to evaluate: ∫ from x=0 to 2 ∫ from y=0 to 2x+3 (3y^2 + 3xy + 2x^2) dy dx.

First, I will perform the inner integral with respect to y, treating x as a constant:
- ∫ 3y^2 dy = y^3
- ∫ 3xy dy = (3x) ∙ (y^2/2) = (3x/2) ∙ y^2
- ∫ 2x^2 dy = 2x^2 ∙ y
Evaluate from y = 0 to y = 2x + 3:
Value = (2x+3)^3 + (3x/2)·(2x+3)^2 + 2x^2·(2x+3).

Next, I expand and simplify this expression in terms of x:
Let A = (2x+3). Then A^2 = 4x^2 + 12x + 9 and A^3 = 8x^3 + 36x^2 + 54x + 27.
- (3x/2)·A^2 = (3x/2)·(4x^2 + 12x + 9) = 6x^3 + 18x^2 + (27/2)x
- 2x^2·A = 2x^2·(2x+3) = 4x^3 + 6x^2
Now sum all three components:
(2x+3)^3 + (3x/2)·(2x+3)^2 + 2x^2·(2x+3) = 18x^3 + 60x^2 + (135/2)x + 27.

Then integrate this result with respect to x from 0 to 2:
- ∫ 18x^3 dx = (18/4)x^4 = (9/2)x^4
- ∫ 60x^2 dx = 60·(x^3/3) = 20x^3
- ∫ (135/2)x dx = (135/4)x^2
- ∫ 27 dx = 27x
Evaluate from x = 0 to x = 2:
(9/2)·(2^4) + 20·(2^3) + (135/4)·(2^2) + 27·(2) = 72 + 160 + 135 + 54 = 421.

Note: The provided data includes an answer field with '421' but there were no answer options listed to analyze. In this analysis, I computed the integral from first principles and confirmed the value 421. Since no alternative choices were given, I cannot compare or explain why other options might be incorrect.

If you expected multiple-choice options, please provide the option list (even partial or reformatted) and I will dissect each choice in turn with reasoning tailored to that set.

MATH0046_24-25 Quiz week 8 (Not assessed)

Question textCalculate [math: ∫02∫02x+33y2+3xy+2x2dydx]\int_{0}^{2}\int_{0}^{2\,x+3} {3\,y^2+3\,x\,y+2\,x^2} \: \mathrm{d}y \mathrm{d}x:[input] Check Question 1

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