Find c>0 such that the area of the region enclosed by the parabolas y = x^2 - c^2 and y = c^2 -x^2 is 230.
c =?
$\displaystyle \int_{-c}^{c}\left[ c^2 - x^2 - \left( x^2 - c^2 \right) \right]dx$
$\displaystyle =2 \int_{0}^{c} \left( 2c^2 - 2x^2 \right) dx$
$\displaystyle = 4 \int_{0}^{c} \left( c^2 - x^2 \right)dx$
$\displaystyle = 4 \left[ c^2 x - \frac {1}{3}x^3 \right]_{0}^{c}$
$\displaystyle = \frac {8}{3}c^3$
check my computation, i was in a rush. i have to leave for a little while
did you see the difference with how i integrated c? i treated it as a constant--which it is
i know, qbkr21 already confirmed it. i believe he has the answer to the question in the back of his text. i hope he gets why i could have integrated between 0 and c though and then multiply by two, furthermore, i hope he gets why i would want to do that in the first place
Note that your integrand is an even function: f(-x) = f(x). In this case
$\displaystyle \int_{-c}^c dx \, f(x) = \int_{-c}^0 dx \, f(x) + \int_0^c dx \, f(x)$
Use the substitution y = -x in the first integral:
$\displaystyle = \int_{c}^0 (-dy) \, f(-y) + \int_0^c dx \, f(x)$
$\displaystyle = -\left ( -\int_0^c dy \, f(-y) \right ) + \int_0^c dx \, f(x)$
But f(-y) = f(y) when f is an even function:
$\displaystyle = \int_0^c dy \, f(y) + \int_0^c dx \, f(x)$
and now just replace the dummy variable y in the first integration with x:
$\displaystyle = \int_0^c dx \, f(x) + \int_0^c dx \, f(x)$
$\displaystyle = 2 \int_0^c dx \, f(x)$
-Dan