Change the order of integration and evaluate?

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Answer 1 · 2018-05-01T08:15:33

Below

If you do change you need to split the integration out as:

$int_0^1 \ int_0^sqrty \ xy \ dx \ dy + int_1^2 \ int_0^(2 - y)\ xy \ dx \ dy$

$= int_0^1 \ (1/2x^2y)_0^sqrty \ dy + int_1^2 \ (1/2x^2y)_0^(2 - y) \ dy$

$=1/2 int_0^1 \ y^2 \ dy + 1/2int_1^2 \ 4y -4y^2 + y^3 \ dy$

$=1/2 ( (y^3/3)_0^1 + (\ 2y^2 -4/3 y^3 + y^4/4 )_1^2) = 3/8$

Whereas:

$int_0^1 \ int_(x^2)^(2-x) \ xy \ dy\ dx$

$= int_0^1 \ ( 1/2 xy^2 )_(x^2)^(2-x)\ dx$

$=1/2 int_0^1 \ 4x - 4x^2 + x^3 - x^5 \ dx$

$=1/2 (2x^2- 4/3 x^3 + x^4/4 - x^6/6)_0^1 = 3/8$

Answer 2 · 2018-05-01T14:31:48

To change the order of integration, please see below.

The region over which we are integrating is bounded by $y = x^2$ and $y = 2-x$ from $x=0$ to $x = 1$

Here is the region:

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To change the order of integration, we need to re-describe the region.

Apparently $y$ goes from $0$ to $2$ and the region is split into two parts.

From $y=0$ to $y=1$ , the bounds are $x=sqrty$ (solve $y = x^2$ for $y$ in the first quadrant) and $x = 0$

From $y=1$ to $y=2$ , the bounds are $x=2-y$ (solve $y = 2-x$ for $y$ ) and $x = 0$

The function to be integrated is $f(x,y) = xy$ so we have

$int_0^1 int_0^(sqrty) xy\ dx\ dy + int_1^2 int_0^(2-y) xy\ dx\ dy$

Please see the other answer for evaluation.