Area under Curve: Find x-intercept & y-intercept

uradnky
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Homework Statement



Find the area bound by y= x - 2\sqrt{x} and y=0


The Attempt at a Solution



X intercept = 4
Y intercept = 0

Using verticle rectangles..

dA = ((Upper curve)-(Lower curve)) dx
\int(dA = \int(0-(x-2\sqrt{x})) dx

A= -(1/2X^2)-((2x^(3/2))/(2/3)) Evaluated from 0 to 4

A= 16

I understand this is an extremely easy question but I just cannot see where I have gone wrong with it. Is there a problem with my signs since the area I am finding is below the x-axis? Sorry if this is a bit messy, any help is appecricated, thanks.
 
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You should be calculating \int2sqrt(x) - x from 0 to 4.
 
That was easy. Thanks.
 
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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