Question

How do I find the area between the curves `y=x^2-4x+3` and `y=3+4x-x^2`?

Answer 1

It helps to first make a scetch of the two curves.
Preferably on the same paper (can't get this done here)

Then find the intersection points where both are equal:
`y=x^2-4x+3=-x^2+4x+3`

You will find `x=0` and `x=4`

Now we need the difference-function (the space between the functions):
`y=(x^2-4x+3)-(-x^2+4x+3)=2x^2-8x`

Then we integrate this function from `0` to `4`

Area = `int_0^4 (2x^2-8x).dx=|_0^4 2/3x^3-4x^2=`

Area = `|(2/3*4^3-4*4^2)-0|=21 1/3`

(we take the absolute, positive value)