# Integral

• April 9th 2009, 12:43 PM
PensFan10
Integral
I can not seem to get the correct answer on the following integral.

Integral from (1-x) up to 2 of (2x+2-y)/(4) dy

Please show the steps so i can see what I am doing wrong. Thanks
• April 9th 2009, 02:25 PM
Jester
Quote:

Originally Posted by PensFan10
I can not seem to get the correct answer on the following integral.

Integral from (1-x) up to 2 of (2x+2-y)/(4) dy

Please show the steps so i can see what I am doing wrong. Thanks

It's interesting that your integrand has both x and y. Is this what you really want

$\frac{1}{4} \int_{1-x}^2 2 x + 2 - y\; dy\;\;?$
• April 9th 2009, 02:26 PM
Jester
Now that's what I call steps (must be a dance teacher) (Rofl)
• April 9th 2009, 02:50 PM
PensFan10
Yes, that is what I want. The answer the book gives is (5x^2+6x-y)/8

It contains both x and y because its a double integral. I cant get the first part.
• April 9th 2009, 03:01 PM
Jester
Quote:

Originally Posted by PensFan10
Yes, that is what I want. The answer the book gives is (5x^2+6x-y)/8

It contains both x and y because its a double integral. I cant get the first part.

Can you give the whole problem (i.e as in the double integral)?
• April 10th 2009, 11:22 AM
PensFan10
Yeah here it is:

i will use S to denote integral

S(from 0 to 1)S(from (1-x) to 2) of (2x+2-y)/(4) dydx
• April 10th 2009, 11:32 AM
Jester
Quote:

Originally Posted by PensFan10
Yeah here it is:

i will use S to denote integral

S(from 0 to 1)S(from (1-x) to 2) of (2x+2-y)/(4) dy dx

$\frac{1}{4}\int_0^1 \int 2x+2-y \,dy\,dx = \frac{1}{4} \int_0^1 2(x+1)y - \left. \frac{y^2}{2} \right|_{1-x}^2 dx$
$= \frac{1}{4} \int_0^1 \left( 4(x+1) - 2\right) - \left( 2(1+x)(1-x) - \frac{(1-x)^2}{2}\right)dx = \frac{1}{8}\int_0^1 5x^2 + 6x + 1\,dx$
Did you get this?
• April 10th 2009, 12:05 PM
PensFan10
Thats the correct answer, but not what i got. Not sure where I was making my mistake. Thanks