Double Integrals - Order of Integration

**billym** · November 24th 2009, 10:05 AM

Find the area of A by evaluating the double integral

$\int\int_A \, dxdy$

where the region A is bounded by the curve $y=x^2-x$ and the line $y=2$

Is this correct:

$\int\int_A \, dxdy = \int_{-1}^{2}\int_{2}^{x^2-x} \, dydx$

Since it says evaluate $\int\int_A \, dxdy$ , do you reckon I have to change the order of integration?

If so, I have:

$\int_{-1/4}^{2}\int_{1/2 \left(1-\sqrt{4y+1} \right)}^{1/2 \left(1+ \sqrt{4y+1} \right)} \, dxdy$

Is this ridiculous?

**tonio** · November 24th 2009, 11:02 AM

Originally Posted by billym

Find the area of A by evaluating the double integral

[tex] $\int\int_A \, dxdy$

where the region A is bounded by the curve $y=x^2-x$ and the line $y=2$

Is this correct:

$\int\int_A \, dxdy = \int_{-1}^{2}\int_{2}^{x^2-x} \, dydx$

Since it says evaluate $\int\int_A \, dxdy$ , do you reckon I have to change the order of integration?

If so, I have:

$\int_{-1/4}^{2}\int_{1/2 \left(1-\sqrt{4y+1} \right)}^{1/2 \left(1+ \sqrt{4y+1} \right)} \, dxdy$

Is this ridiculous?

It looks pretty absurd: better, just change $dxdy$ by $dydx$ and do the first, much easier, integration.

Tonio

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