Question

Evaluate Integral 0 power 3 integral 1 power 2 xy(1+x+y)dxdy

Answer 1

$\displaystyle\,\\I=\int_0^3\int_{1}^2xy(1+x+y)~dx~dy\\ \\I=\int_0^3\left(\int_{1}^2xy+x^2y+xy^2~dx\right)~dy\\\\I=\int_0^3\left.\left(\dfrac{x^2y}{2}+\dfrac{x^3y}{3}+\dfrac{x^2y^2}{2}\right)\right|_{x=1}^{x=2}~dy\\\\I=\int_0^3\dfrac{3y}{2}+\dfrac{7y}{3}+\dfrac{3y^2}{2}~dy\\\\I=\left.\left(\dfrac{23y^2}{12}+\dfrac{y^3}{2}\right)\right|_{y=0}^{y=3}\\\\I=\dfrac{69}{4}+\dfrac{27}{2}\\ \\\\\boxed{I=\dfrac{123}{4}}$

Answer 2

Answer:

$\int\limits^3_0 \int\limits^2_1 { xy(1+x+y)dxdy} \,=\frac{123}{4}$

Step-by-step explanation:

Given the double integral,

$I=\int\limits^3_0 \int\limits^2_1 { xy(1+x+y)dxdy} \,$

$=\int\limits^3_0 \int\limits^2_1 {( xy+x^{2}y +xy^{2} )dxdy}$

$=\int\limits^3_0\Big( \int\limits^2_1 {( xy+x^{2}y +xy^{2} )dx\Big)dy}$

$=\int\limits^3_0\Big( \Big| \frac{x^{2} y}{2} +\frac{x^{3} y}{3} +\frac{x^{2} y^{2} }{2} \Big|\limits^2_1\Big)dy}$

$=\int\limits^3_0\Big( \frac{(2^{2}-1) y}{2} +\frac{(2^{3}-1) y}{3} +\frac{(2^{2}-1) y^{2} }{2} \Big)dy}$

$=\int\limits^3_0\Big( \frac{3 y}{2} +\frac{7y}{3} +\frac{3 y^{2} }{2} \Big)dy}$

$=\Big| \frac{3 y^{2} }{4} +\frac{7y^{2} }{6} +\frac{3 y^{3} }{6} \Big|\limits^3_0$

$= \frac{3 \times 3^{2} }{4} +\frac{7\times 3^{2} }{6} +\frac{3\times 3^{3} }{6}$

$= \frac{ 27 }{4} +\frac{21}{2} +\frac{27}{2}$

$= \frac{ 27 }{4} +24$

$= \frac{ 27+96 }{4}$

$=\frac{ 123 }{4}$

Therefore, $\int\limits^3_0 \int\limits^2_1 { xy(1+x+y)dxdy} \,=\frac{123}{4}$