#1**+1 **

Evaluate the area bounded by the curves y=x²-4x+3 and y = -x²+2x+3

I have graphed it but this is not strictly necessary.

You should have a basic understanding of the shape of the graph though and you need the x values for points of intescetion

\(area = \displaystyle \int_0^3 [-x^2+2x+3-(x^2-4x+3)]\;dx\\ area = \displaystyle \int_0^3 [-2x^2+6x]\;dx\\ area = \left [ \frac{-2x^3}{3}+3x^2 \right ]_0^3\\~\\ area = -2*9+27 \\~\\ area=9\;u^2 \)

I have also looked at the graph to see if this passes a reasonable test. I think that it does.

LaTex

area = \displaystyle \int_0^3 [-x^2+2x+3-(x^2-4x+3)]\;dx\\

area = \displaystyle \int_0^3 [-2x^2+6x]\;dx\\

area = \left [ \frac{-2x^3}{3}+3x^2 \right ]_0^3\\~\\

area = -2*9+27 \\~\\

area=9\;u^2

Melody Jul 21, 2021