Question

Evaluate the double integral. ∫∫D (2x+y) dA, D = 1 ≤ y ≤ 4, y − 3 ≤ x ≤ 3

Answer 1

The double integral ∫∫_D (2x+y) dA over the domain D is solved using iterated integrals, starting with integration with respect to x and followed by y, using the limits provided by the domain's inequalities.

Step-by-step explanation:

The student has asked to evaluate the double integral of the function 2x+y over the domain D, given by the inequalities 1 ≤ y ≤ 4 and y − 3 ≤ x ≤ 3.

To solve this, we will apply the iterated integral technique, integrating with respect to x first, and then with respect to y.

First, we set up the integral with respect to x: ∫ (2x+y) dx from x = y - 3 to x = 3.

Next, we compute the antiderivative with respect to x, which will be x² + yx, evaluated from y - 3 to 3.

After simplifying the expression, we then set up the integral with respect to y from 1 to 4.

We compute the antiderivative with respect to y and then evaluate it at the limits of integration to obtain the final result.

Answer 2

The value of the double integral is 145.5.

Here's the evaluation of the double integral ∫∫D (2x+y) dA, D = (x, y) :

1. Set up the iterated integral:

We'll integrate first with respect to x, then with respect to y, based on the given bounds for D:

∫₁⁴ ∫y-3³ (2x+y) dx dy

2. Evaluate the inner integral:

∫₁⁴ [x² + xy]y-3³ dy

∫₁⁴ (9 + 3y - (y² - 6y + 9)) dy

∫₁⁴ (3y² + 9y) dy

3. Evaluate the outer integral:

[y³ + (9/2)y²]₁⁴

(64 + 81) - (1 + (9/2))

= 145.5

Therefore, the value of the double integral is 145.5.