1. ## 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

2. 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

$\displaystyle \frac{1}{4} \int_{1-x}^2 2 x + 2 - y\; dy\;\;?$

3. Now that's what I call steps (must be a dance teacher)

4. 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.

5. 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)?

6. 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

7. 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
$\displaystyle \frac{1}{4}\int_0^1 \int 2x+2-y \,dy\,dx = \frac{1}{4} \int_0^1 2(x+1)y - \left. \frac{y^2}{2} \right|_{1-x}^2 dx$
$\displaystyle = \frac{1}{4} \int_0^1 \left( 4(x+1) - 2\right) - \left( 2(1+x)(1-x) - \frac{(1-x)^2}{2}\right)dx = \frac{1}{8}\int_0^1 5x^2 + 6x + 1\,dx$
Did you get this?

8. Thats the correct answer, but not what i got. Not sure where I was making my mistake. Thanks