Evaluate the double integral int^(1_0) int^(0_-2) (2x+3y) dy dxcan you explain Thank u

Expert Answers
jeew-m eNotes educator| Certified Educator

This double integral should have to be done in two parts. first we have to integrate with respect to (w.r.t) y and then w.r.t x. When integrating w.r.t y then x becomes constant and vice versa.

 

`int^1_0 int^0_-2 (2x+3y) dy dx`

= `int^1_0 [2xy+3y^2/2]_(-2)^0 dx`    -----> integration w.r.t. y. x is constant

= `int^1_0[2x(-2)+3(-2)^2/2-0]dx`  -----> apply limits of y

= `int^1_0 [-4x+6] dx`

= `[-2x^2+6x]_0^1`

= [-2+6-0]

= 4

 

`So` `int^1_0 int^0_-2 (2x+3y) dy dx``= 4`

lfryerda eNotes educator| Certified Educator

This is an iterated integral, which means that you evaluate the dy integral first and then the dx integral second.  In each case, you hold the other variable as a constant.

`int_0^1 int_{-2}^0 (2x+3y)dy dx`    integrate y

`=int_0^1(2xy+3/2y^2)_{y=-2}^{y=0}dx`   simplify

`=int_0^1(-4x+6)dx`

`=(-4/2x^2+6x)_0^1`

`=-2+6`

`=4`

The double integral is 4.