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How would I set up the folowing double integral

$4\displaystyle \iint x^{1/2}dA$$

where the domain is the closed region bounded by $y=x$ and $y=x^2-3x$

I know the domain is curve shaped so I would do dx cut vertically

So would my integral be

$$\int_0^4\int_{x^2-3x}^x dydx$$

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    $\begingroup$ Now the limits are OK, but you forgot the integrand. $\endgroup$
    – Git Gud
    Apr 25, 2014 at 15:00
  • $\begingroup$ I see that makes sense $\endgroup$ Apr 25, 2014 at 15:02
  • $\begingroup$ It would be good if you could post an answer to the question yourself so this doesn't come up as unanswered. $\endgroup$
    – Git Gud
    Apr 25, 2014 at 16:15
  • $\begingroup$ ok I shall do it $\endgroup$ Apr 25, 2014 at 17:52

1 Answer 1

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The way it would be set up is

$\int_0^4\int_{y=x^2-3x}^{y=x}(x)^{1/2}dydx$

taking the first integral would give

$(x)^{3/2}-x^{5/2}+3x^{3/2}$

$\int_0^4 4x^{3/2}-x^{5/2}dx$

$\frac{8}{5}x^{5/2}-\frac{2}{7}x^{7/2}$

=$\frac{8}{5}(4)^{5/2}-\frac{2}{7}(4)^{7/2}-0$

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