
double integral problem
Use double integral to find the area bounded by the equation:
y= x(x2)(x+1) , y=0
I have attempted it by doing the following:
For x fixed, y varies from y= x(x2)(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)

Area
Hello,
If you draw the graph you'll notice that you have two areas. One extending from $\displaystyle x_{1}=1$ to $\displaystyle x_{2}=0$ and the other area extending from $\displaystyle x_{2}=0$ to $\displaystyle x_{3}=2$. So correct the limits.
Furthermore make sure to consider the sign of the integrals (one area is above $\displaystyle y=0$ and the other is under $\displaystyle y=0$ ).

I have drawn the curve and i am integrating for
while x is fixed
y varies from y=0 to y= x(x2)(x+1), x varies from x=2 to x=0
+
while x is fixed
y varies from y= x(x2)(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
help please!

Then show what you have done. What did you get for each part of
$\displaystyle \int_{1}^0 (x^3 x^2 2x) dx \int_0^2(x^3 x^2 2x)dx$?