# double integral problem

• May 4th 2010, 08:24 AM
mehn
double integral problem
Use double integral to find the area bounded by the equation:
y= x(x-2)(x+1) , y=0

I have attempted it by doing the following:
For x fixed, y varies from y= x(x-2)(x+1) to y=0
x varies from x=2 to x=-1
But while solving the integral i'm not reaching the required answer.

Can anyone help me out please?
I have exams this Friday!
Thanks bunches (Crying)
• May 4th 2010, 11:23 AM
surjective
Area
Hello,

If you draw the graph you'll notice that you have two areas. One extending from $x_{1}=-1$ to $x_{2}=0$ and the other area extending from $x_{2}=0$ to $x_{3}=2$. So correct the limits.

Furthermore make sure to consider the sign of the integrals (one area is above $y=0$ and the other is under $y=0$ ).
• May 4th 2010, 09:43 PM
mehn
I have drawn the curve and i am integrating for
while x is fixed
y varies from y=0 to y= -x(x-2)(x+1), x varies from x=2 to x=0
+
while x is fixed
y varies from y= x(x-2)(x+1) to y=0, x varies from x=0 to x=-1

but still i'm not reaching the answer.
I'm always getting -2.25 as answer when it should have been 37/12

$\int_{-1}^0 (x^3- x^2- 2x) dx- \int_0^2(x^3- x^2- 2x)dx$?