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

Jan 29, 2015

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:
$y = {x}^{2} - 4 x + 3 = - {x}^{2} + 4 x + 3$

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

Now we need the difference-function (the space between the functions):
$y = \left({x}^{2} - 4 x + 3\right) - \left(- {x}^{2} + 4 x + 3\right) = 2 {x}^{2} - 8 x$

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

Area = ${\int}_{0}^{4} \left(2 {x}^{2} - 8 x\right) . \mathrm{dx} = {|}_{0}^{4} \frac{2}{3} {x}^{3} - 4 {x}^{2} =$

Area = $| \left(\frac{2}{3} \cdot {4}^{3} - 4 \cdot {4}^{2}\right) - 0 | = 21 \frac{1}{3}$

(we take the absolute, positive value)