Let R be the rectangle consisting of all points (x, y), such that 0≤x≤4,1≤y≤3, and calculate the double integral. ∫R∫(2x+3y)dxdy

Step 1 Given integral is: ∫R∫(2x+3y)dxdy where R be the rectangle consisting of all points (x, y), such that 0≤x≤4,1≤y≤3 Step 2 Then the integral becomes: I=∫13∫04(2x+3y)dxdy I=∫13[2(x22)+3xy]04dy I=∫13[x2+3xy]04dy I=∫13[42+3(4)y−0−0]dy I=∫13(16+12y)dy I=[16y+12(y22)]13 I=[16y+6y2]13 I=[16(3)+6(3)2−16(1)−6(1)2] I=(48+54-16-6) I=80 Hence the value of the double integral is 80.

Explained: "Let R be the rectangle consisting of all..."

Burhan Hopper

Answered question

2021-10-25

Let R be the rectangle consisting of all points (x, y), such that

0 \leq x \leq 4, 1 \leq y \leq 3

, and calculate the double integral.

\int_{R} \int (2 x + 3 y) d x d y

Answer & Explanation

svartmaleJ

Skilled2021-10-26Added 92 answers

Step 1
Given integral is:

\int_{R} \int (2 x + 3 y) d x d y

where R be the rectangle consisting of all points (x, y), such that

0 \leq x \leq 4, 1 \leq y \leq 3

Step 2
Then the integral becomes:

I = \int_{1}^{3} \int_{0}^{4} (2 x + 3 y) d x d y

I = \int_{1}^{3} {[2 (\frac{x^{2}}{2}) + 3 x y]}_{0}^{4} d y

I = \int_{1}^{3} {[x^{2} + 3 x y]}_{0}^{4} d y

I = \int_{1}^{3} [4^{2} + 3 (4) y - 0 - 0] d y

I = \int_{1}^{3} (16 + 12 y) d y

I = {[16 y + 12 (\frac{y^{2}}{2})]}_{1}^{3}

I = {[16 y + 6 y^{2}]}_{1}^{3}

I = [16 (3) + 6 {(3)}^{2} - 16 (1) - 6 {(1)}^{2}]

I=(48+54-16-6)
I=80
Hence the value of the double integral is 80.

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