Question

eevalute double integral

Answer 1

Topic :-

Double Integral

To Integrate :-

$\displaystyle \int_{0}^{3} \int_{0}^{2} xy(x+y)\cdot dx\cdot dy$

Solution :-

$\displaystyle \int_{0}^{3} \int_{0}^{2} xy(x+y)\cdot dx\cdot dy$

$\displaystyle \int_{0}^{3} \int_{0}^{2} (x^2y+xy^2)\cdot dx\cdot dy$

Solving first integral,

'y' will act as constant while solving first integral as function is being integrated with respect to 'dx'.

$\displaystyle \int_{0}^{3} \left( \int_{0}^{2} (x^2y+xy^2)\cdot dx \right) dy$

$\displaystyle \int_{0}^{3} \left[ y\dfrac{x^{2+1}}{2+1}+y^2\dfrac{x^{1+1}}{1+1} +C\right]_{0}^{2} dy$

$\displaystyle \int_{0}^{3} \left[ y\dfrac{x^3}{3}+y^2\dfrac{x^2}{2} +C\right]_{0}^{2} dy$

$\left ( \because \displaystyle \int kx^n\cdot dx=k\dfrac{x^{n+1}}{n+1}+C\right)$

$\displaystyle \int_{0}^{3} \left[ y\dfrac{2^3}{3}+y^2\dfrac{2^2}{2} +C-y\dfrac{0^3}{3}-y^2\dfrac{0^2}{2} -C\right] dy$

$\displaystyle \int_{0}^{3} \left[ y\dfrac{2^3}{3}+y^2\dfrac{2^2}{2} +\not{C}-0-0-\not{C}\right] dy$

$\displaystyle \int_{0}^{3} \left[ \dfrac{8y}{3}+2y^2 \right] dy$

Solving second integral,

$\displaystyle \int_{0}^{3} \left( \dfrac{8y}{3}+2y^2 \right) dy$

$\left[ \dfrac{8y^{1+1}}{3(1+1)}+2\dfrac{y^{2+1}}{2+1} +C'\right]_{0}^{3}$

$\left[ \dfrac{8y^{2}}{3(2)}+\dfrac{2y^{3}}{3} +C'\right]_{0}^{3}$

$\left ( \because \displaystyle \int ky^n\cdot dy=k\dfrac{y^{n+1}}{n+1}+C'\right)$

$\left[ \dfrac{4(3^{2})}{3}+\dfrac{2(3^{3})}{3} +C'-\dfrac{4(0^{2})}{3}-\dfrac{2(0^{3})}{3} -C'\right]$

$\left[ \dfrac{4(9)}{3}+\dfrac{2(27)}{3} +\not{C'}-0-0-\not{C'}\right]$

$\left[ \dfrac{36}{3}+\dfrac{54}{3} \right]$

$12+18$

$30$

Answer :-

So, value of given double integral is 30.