Find the areas of the regions whose boundaries are given

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I have three questions:

Homework Statement



Find the areas of the regions whose boundaries are given.

Homework Equations



y=x^3-3
y=1

The Attempt at a Solution



x=-2, x=2
I got -10.67 but I know this can't be true because you can't have a negative area.

Homework Statement



Find the areas of the regions whose boundaries are given.

Homework Equations



y^2=x
x+y=2

The Attempt at a Solution



y=1, y=-2
I got -4.5, but again that can't be right because it's negative :(

Homework Statement



Find the areas of the regions whose boundaries are given.

Homework Equations



y=x^3-2x^2-3x
y=0

The Attempt at a Solution



I got to x=0, x=-1, and x=3 but I don't know where to go from here.

Thanks for any help! :)
 
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Let's just start with the first one. How did you get x=-2, x=2 or was that given? The curves y=x^3-3 and y=1 don't enclose any bounded region.
 
many apologies... it should have been x^2-3
 
If you get a negative area then you have the two functions in the wrong order. For x between -2 and 2, the graph of x2- 3 is below the graph of y= 1. You should be integrating \int [1- (x^2-3)]dx= \int (4- x^2)dx. That, integrated between -2 and 2, is positive.
 
And for the last problem, you would have two different enclosed areas. One would be between -1 and 0 and the other between 0 and 3.
This means you would have to set up 2 different equations and find the sum of the areas.
HINT: The equations are very similar just siwtched in order.
 
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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