Question

Calculate the double integral. $\iint_{R}{\color{red}4} xye^{x^{2}y}\hspace*{3pt}dA, \quad R = [0, 1] \times [0, {\color{red}7}] $

Answer 1

Answer:

$\mathbf{\int \int _R \ 4xy e^{x^2 \ y} \ dA = 2 (e^7 -8)}$

Step-by-step explanation:

Given that:

$\int \int _R 4xye^{x^2 \ y} \ dA, R = [0,1]\times [0,7]$

The rectangle R = [0,1] × [0,7]

R = { (x,y): x ∈ [0,1] and y ∈ [0,7] }

R = { (x,y): 0 ≤ x ≤ 1 and 0 ≤ x ≤ 7 }

$\int \int _R \ 4xy e^{x^2 \ y} \ dA = \int^{7}_{0}\int^{1}_{0} 4xye^{x^2 \ y} \ dx dy$

$\int \int _R \ 4xy e^{x^2 \ y} \ dA = \int^{7}_{0} \begin {bmatrix} ye^{yx^2} \dfrac{4}{2y} \end {bmatrix}^1 _ 0 \ dy$

$\int \int _R \ 4xy e^{x^2 \ y} \ dA = \int^{7}_{0} \begin {bmatrix} ye^{y1^2} \dfrac{4}{2y} - ye^{y0^2} \dfrac{4}{2y} \end {bmatrix}\ dy$

$\int \int _R \ 4xy e^{x^2 \ y} \ dA = \int^{7}_{0} \dfrac{4}{2}(e^y -1) \ dy$

$\int \int _R \ 4xy e^{x^2 \ y} \ dA = \dfrac{4}{2}[e^y -1]^7_0 \ dy$

$\int \int _R \ 4xy e^{x^2 \ y} \ dA = 2 [(e^7 -7)-(e^0 -0)]$

$\int \int _R \ 4xy e^{x^2 \ y} \ dA = 2 [(e^7 -7)-1]$

$\mathbf{\int \int _R \ 4xy e^{x^2 \ y} \ dA = 2 (e^7 -8)}$