# Integral and area

1. May 19, 2006

### merced

Find the area of the region

y = x
x + 2y = 0
2x + y = 3

http://img149.imageshack.us/img149/4274/math37yo.th.jpg [Broken]

So, I decided to integrate with respect to y.

And got the integral from -1 to 0 of [(3-y)/2 - (-2y)]dy + the integral from 0 to 1 of [(3-y)/2 - y]dy.

Integrating that, I got

[3y/2 + 3y^2/4]0-1 + [3y/2 - 3y^2/4]10 = 0

The book integrated with respect to x, but I wanted to do it in terms of y.
Is it possible integrating it with respect to y?

Last edited by a moderator: May 2, 2017
2. May 19, 2006

### HallsofIvy

Okay, the lines x+ 2y= 0 and 2x+y= 3 intersect at (3, -3/2) (the lowest point on your graph) while 2x+y= 3 and y= x intersect at (1, 1). At y= 0, of course, the "left edge" changes formula. Integrate (3- y)/2 - (-2y) from -3/2 to 0 and then integrate y- (-2y) from 0 to 1.

3. May 19, 2006

### merced

"Okay, the lines x+ 2y= 0 and 2x+y= 3 intersect at (3, -3/2) (the lowest point on your graph) while 2x+y= 3 and y= x intersect at (1, 1). At y= 0, of course, the "left edge" changes formula."Integrate (3- y)/2 - (-2y) from -3/2 to 0 and then integrate y- (-2y) from 0 to 1."

Actually, they intersect at (2, -1).
That's why I was integrating [(3- y)/2 - (-2y)] from -1 to 0
(at x = 0, the upper function changes from y = x to 2x + y = 3.

"Integrate (3- y)/2 - (-2y) from -3/2 to 0 and then integrate y- (-2y) from 0 to 1."

Do you mean integrate [(3-y)/2 - y] from 0 to 1?
I don't understand where you're getting y - (- 2y) from.

4. May 20, 2006