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

Asked on by berry90

jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted on

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 | High School Teacher | (Level 2) Educator

Posted on

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.

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