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

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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`

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.**