Question

Evaluate int int x int(5-2x-y)dx dy
where A is given by y = 0, x + 2y = 3, x = y ^ 2

Answer 1

Topic:

Integration

Solution:
We need to evaluate the following integral.

$\displaystyle\rm I =\iint\limits_A (5 - 2x - y) \ dx\ dy$

$\mathrm{where\ A\ is\ given\ by\ y = 0, x + 2y = 3\ and\ x =y^2.}$

• First of all plot the graphs for given equations and find the limits for the integral as the region A is given by provided equations.

Interpreting region A, we get the following limits:
${\displaystyle\rm I =\int_0^1\int_{y^2}^{3 - 2y} (5 - 2x - y) \ dx\ dy}$

Solving the inner integral, w.r.t. x keeping y's as constant.

${\displaystyle\rm I =\int_0^1 (5x - x^2 - yx) \bigg|^{3-2y}_{y^2}\ dy}$

${\displaystyle\rm I =\int_0^1 \Big(5(3-2y) - (3-2y)^2 - y(3-2y)\Big) - \Big(5(y^2) - (y^2)^2 - y(y^2)\Big) dy}$

${\displaystyle\rm I =\int_0^115 - 10y- (9 + 4y^2 - 12y)- 3y +2y^2 - 5y^2 + y^4 + y^3\ dy}$

${\displaystyle\rm I =\int_0^115 - 10y- 9 - 4y^2 + 12y- 3y +2y^2 - 5y^2 + y^4 + y^3\ dy}$

${\displaystyle\rm I =\int_0^1(y^4 + y^3 - 7 y^2 - y + 6)\ dy}$

Solving the integral w.r.t. y

${\displaystyle\rm I =\dfrac{y^5}{5} + \dfrac{y^4}{4} - \dfrac{7y^3}{3} - \dfrac{y^2}{2} + 6y\bigg|^1_0}$

${\displaystyle\rm I =\dfrac{1^5}{5} + \dfrac{1^4}{4} - \dfrac{7(1)^3}{3} - \dfrac{(1)^2}{2} + 6(1)}$

${\displaystyle\rm I =\dfrac{1}{5} + \dfrac{1}{4} - \dfrac{7}{3} - \dfrac{1}{2} + 6}$

${\displaystyle\rm I =\dfrac{217}{60}}$

${\displaystyle\rm I \approx 3.616667}$

So the required answer is,

$\blue{\underline{\boxed{\sf\iint_A\ \ (5 - 2x - y) \ dx\ dy = \dfrac{217}{60}}}}$