1. ## Area between curves

1) $y = |x|$ and $y = x^2-6$

Using the definition of absolute value of x as a piecewise function, the intersects are -3 and 3.

$\int^3_{-3} (|x| - x^2-6)dx = \int^0_{-3}(-x-x^2-6)dx + \int^3_0 (x-x^2-6)dx$

$\frac{-x^2}{2}-\frac{x^3}{3}-6x |^0_{-3} + \frac{x^2}{2} -\frac{x^3}{3}-6x|^3_0$

$0 - (\frac{9}{2} - \frac{27}{3}-18)+(\frac{9}{2}-\frac{27}{3}-18)-0$

$0 -(\frac{-45}{2})+(\frac{-45}{2})$
But that can't be right....

2) $y = 7x^2$ and $y = \frac{1}{7}x$

This one I think i could do if I could get started. But I keep end up having to find the square root of 1/7, which I can't seem to figure out how to express as a fraction. Maybe just a may problem, but I don't know.

2. Originally Posted by Open that Hampster!
1) $y = |x|$ and $y = x^2-6$

Using the definition of absolute value of x as a piecewise function, the intersects are -3 and 3.

$\int^3_{-3} (|x| - x^2-6)dx = \int^0_{-3}(-x-x^2-6)dx + \int^3_0 (x-x^2-6)dx$

$\frac{-x^2}{2}-\frac{x^3}{3}-6x |^0_{-3} + \frac{x^2}{2} -\frac{x^3}{3}-6x|^3_0$

$0 - (\frac{9}{2} - \frac{27}{3}-18)+(\frac{9}{2}-\frac{27}{3}-18)-0$

$0 -(\frac{-45}{2})+(\frac{-45}{2})$
But that can't be right....

2) $y = 7x^2$ and $y = \frac{1}{7}x$

This one I think i could do if I could get started. But I keep end up having to find the square root of 1/7, which I can't seem to figure out how to express as a fraction. Maybe just a may problem, but I don't know.

1. take advantage of the symmetry of the two graphs ...

$2 \int_0^3 x - (x^2-6) \, dx = 27
$
... fixed error

2. find the points of intersection ...

$7x^2 = \frac{x}{7}$

$49x^2 = x$

$49x^2 - x = 0$

$x(49x-1) = 0$

$x = 0$ , $x = \frac{1}{49}$

$\int_0^{\frac{1}{49}} \frac{x}{7} - 7x^2 \, dx$

3. Originally Posted by skeeter
1. take advantage of the symmetry of the two graphs ...

$2 \int_0^3 x - (x^2-6) \, dx = 54
$

2. find the points of intersection ...

$7x^2 = \frac{x}{7}$

$49x^2 = x$

$49x^2 - x = 0$

$x(49x-1) = 0$

$x = 0$ , $x = \frac{1}{49}$

$\int_0^{\frac{1}{49}} \frac{x}{7} - 7x^2 \, dx$

as usual u nailed it Skeetor, but the first one evaluates to 27 not 54.