Quick question about total area (involving integrals)

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Homework Statement



Find the total area betwen the curve y=x^2-4 and the interval [0,3].

Homework Equations





The Attempt at a Solution



Okay, so I already know the answer, but I'm unsure about one thing.

It's ture that you have to break the equation such that it says \int_{2}^{3} x^2 dx - \int_{0}^{2} 4 dx, but how would you go about finding which upper and lower limits to set for the two integration problems like that?

I know how to take care of the rest once I find those, but I'm not sure how to get the upper and lower limits once I break the equation into two.
 
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The integral will give you the area above the x-axis minus the area below the x-axis. If you want the sum of the two instead of the difference then you need to figure out where the curve crosses the x-axis.
 
And you can't break up the integral like you did. You can either do
\int^3_0 x^2 - 4 dx= \int^3_0 x^2 dx - \int^3_0 4 dx or
\int^3_0 x^2-4 dx = \int^3_2 x^2 - 4 dx + \int^2_0 x^2 - 4 dx
 
Aerosion said:

Homework Statement



Find the total area betwen the curve y=x^2-4 and the interval [0,3].
Do you mean "find the area between the curve y= x2- 4 and the x-axis for x in [0,3]?

Homework Equations





The Attempt at a Solution



Okay, so I already know the answer, but I'm unsure about one thing.

It's ture that you have to break the equation such that it says \int_{2}^{3} x^2 dx - \int_{0}^{2} 4 dx, but how would you go about finding which upper and lower limits to set for the two integration problems like that?

I know how to take care of the rest once I find those, but I'm not sure how to get the upper and lower limits once I break the equation into two.

Homework Statement

No. y= x2- 4 crosses the x-axis at x= 2. For x> 2, y is negative and the integral is the negative of the area. The area between the curve and the x-axis for x between 2 and 3 is
-\int_2^3 (x^2- 4)dx
so the "total area" between the curve and the x-axis is
\int_0^2 (x^2- 4)dx- \int_2^3 (x^2- 4)dx



Homework Equations





The Attempt at a Solution

 
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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