Thread: Double integral - absolute values

1. Double integral - absolute values

$\int_{0}^{1} \int_{0}^{1} |x - y| dydx$

- x + y if y < x
x - y if y >= x

so i used this:
$\int_{0}^{1}(\int_{x}^{1} (x - y) + \int_{0}^{x}(-x +y) dy)dx$

$\int_{0}^{1} [xy - \frac{y^2}{2}]_{x}^{1} + [-xy + \frac{y^2}{2}]_{0}^{x}dx$
$\int_{0}^{1} x - \frac{1}{2} - x^2 + \frac{x^2}{2} + (-x^2 + \frac{x^2}{2} + 0 + 0 )dx$

$\int_{0}^{1} x - \frac{1}{2} - x^2dx$

in solving this:
i got a result of - 1/3

but the answer at back of book is 1/3

2. Remember, Adam, absolute values are always positive.

3. x - y for y in [0,x]
x is the maximum value
x is greater than or equal to y
x - y >= 0 so |x-y| = x-y

x - y for y in [x,1]
x is the minimum value
x is less than or equal to y
x - y <= 0 so |x-y| = -x+y

It looks to me like you just have it backwards. Give it another go.

4. back of book is wrong:

My prof told me to integrate inside 1st:

$\int_{0}^{1} \int_{0}^{1} |x - y| dy dx$

$\int_{0}^{1} |xy - \frac{y^2}{2}| dx$

$\int_{0}^{1} |x - \frac{1}{2} |dx$
....
....
....

then the answer is 1/4 solving this

so is this method correct?

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double integration of mod(x (plus)y)

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