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# area bounded by curves

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

Jul 21, 2021

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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

Jul 21, 2021