Question

Find an approximate value of the double integral below where 𝑅 is the rectangular region having

vertices (−1, 1) and (2, 3). Take a partition of 𝑅 formed by the lines 𝑥 = 0, 𝑥 = 1, and 𝑦 = 2, and take (𝑢𝑖

, 𝑣𝑖) at the

center of the 𝑖th sub region.

∬(3𝑦 − 2𝑥

2)𝑑𝐴

𝑅

Answer 1

Question: Find an approximate value of the double integral below where 𝑅 is the rectangular region having

vertices (−1, 1) and (2, 3). Take a partition of 𝑅 formed by the lines 𝑥 = 0, 𝑥 = 1, and 𝑦 = 2, and take (𝑢𝑖, 𝑣𝑖) at the center of the 𝑖th sub region.

∬(3𝑦 − 2𝑥^2)𝑑𝐴

𝑅

Solution:

$\int\int_R(3y-2x^2)dA \\=\int_{x=-1}^{x=2}\int_{y=1}^{y=3}(3y-2x^2)dxdy \\=\int_{x=-1}^{x=2}(\dfrac{3y^2}2-2x^2y)|_{y=1}^{y=3}dx \\=\int_{x=-1}^{x=2}[(\dfrac{3(3)^2}2-2x^2(3))-(\dfrac{3(1)^2}2-2x^2(1))]dx \\=\int_{x=-1}^{x=2}[(\dfrac{27}2-6x^2)-(\dfrac{3}2-2x^2)]dx \\=\int_{x=-1}^{x=2}(12-4x^2)dx \\=(12x-\dfrac{4x^3}3)|_{x=-1}^{x=2} \\=(12(2)-\dfrac{4(2)^3}3)-(12(-1)-\dfrac{4(-1)^3}3) \\=(24-\dfrac{32}3)-(-12+\dfrac{4}3) \\=36-12 \\=24\ sq.\ units$