Question

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

Answer 1

Hello !

Answer is `64/3`.

First, graph the curves. You get

Second, calculate the abscises of intersection : solve
`x^2-4x+3 = 3+4x-x^2`,
so
`2x^2-8x = 0` or `2x(x-4)=0`
then `x = 0` or `x=4`.

Finally, calculate the area `\mathcal{A}` with an integral :
`\mathcal{A}= int_0^4 [(3+4x-x^2) - (x^2-4x+3) ]dx = int_0^4 (-2x^2+8x)dx = [-2/3 x^3 + 4x^2]_0^4`

Then `\mathcal{A} = -2/3 \times 4^3 - 4^3 = 4^3/3 = 64/3`.