Double integration over a region

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Homework Statement



∫∫ (2x+4y+1) dA, R: y = x^{2}, y = x^{3}

Homework Equations



R∫ f(x,y) dA

The Attempt at a Solution


Since there was no direct question, I assumed to just evaluate with whatever I was given.

I would really like to make sure that my integrals are set up correctly:

∫^{1}_{0} ∫^{x^{2}}_{x^{3}} (2x+4y+1) dydx
 
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aaronfue said:

Homework Statement



∫∫ (2x+4y+1) dA, R: y = x^{2}, y = x^{3}

Homework Equations



R∫ f(x,y) dA

The Attempt at a Solution


Since there was no direct question, I assumed to just evaluate with whatever I was given.

I would really like to make sure that my integrals are set up correctly:

∫^{1}_{0} ∫^{x^{2}}_{x^{3}} (2x+4y+1) dydx

EDIT : Wow read that wrong I am half asleep. Sorry. Looks good.
 
Last edited:
LOL...great! I was reading all through my text and ready to call it quits! Thank you Zondrina.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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