Calculate the double integral

Question

∫ ∫ 𝐑 ( 2 𝑥 + 8 𝑦 + 16 ) 𝑑 𝐴

where 𝐑 is the region: 0 ≤ 𝑥 ≤ 4 , 0 ≤ 𝑦 ≤ 1.

Answer 1

So first off, we can see x ranges from 0 to 4, so our integration symbol for x will be ∫₀⁴. For y, we can see it ranges from 0 to 1, so our integration symbol for y will be ∫₀¹.

Since this is a double integral and since our region consists of x and y, we will be integrating with respect to x once and integrating with respect to y once, so our setup will be the following: ∫₀¹∫₀⁴(2x + 8y + 16)dxdy. We will be integrating with respect to x first. It will also work to integrate with respect to y first.

We will be integrating from inside out. So:

∫₀¹[(2x^2)/2 + (8x^1y)/1 + 16x^1]₀⁴dy = ∫₀¹[x^2 + 8xy + 16x]₀⁴dy

∫₀¹[[4^2 + (8)(4)y + 16(4)] - [0^2 + (8)(0)y + 16(0)]]dy = ∫₀¹[80 + 32y - 0]dy = ∫₀¹[80 + 32y]dy

Now y,

[(80y)/1 + (32y^2)/2]₀¹ = [80y + 16y^2]₀¹

[[(80)(1) + (16)1^2] - [(80)(0) + (16)0^2]] = 96 - 0 = 96

So, the answer is 96.

Answer 2

Hi Jetpackdan!

The region of integration is just a rectangle, so we can set this up as an iterated integral as follows.

∫₀¹∫₀⁴ (2x + 8y + 16) dxdy

x-integration:

∫₀¹ (x² + 8xy + 16x)₀⁴ dxdy

∫₀¹ (16 + 32y + 64) dxdy

∫₀¹ (32y + 80) dxdy

y-integration:

(16y² + 80y)₀¹

16 + 80

96

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