<s>[INST] Region(area) enclosed between line and parabola The problem is to find the area of the region enclosed by the given parabola and line:
$y=-x^2+3x-1$
$y=x-2$
I tried solving the definite integral, but the roots (points of intersection of the parabola and line) on the graph were too weird and I ended up with the wrong solution. Does anyone know a more efficient method?
 [/INST] This is a method where you do not need to find the actual roots from the graph, from which the problem of "weird points of intersection" will be completely eradicated. Trust me.
From $-x^2+3x-1=x-2$, $x^2-2x-1=0$.
Letting $a$ and $b$ (where $a<b$) be the solutions of $x^2-2x-1=0$, $a+b=-2$ and $ab=1$, due to the root-coefficient relationship.
Therefore, the area, $S = \int_a^b\left[-x^2+3x-1-\left(x-2\right)\right]dx = -\int_a^b\left(x^2-2x-1\right)dx = -\int_a^b\left(x-a\right)\left(x-b\right)dx = \frac{1}{6}\left(b-a\right)^3$
From $(b-a)^2=(a+b)^2-4ab=8$, $b-a=2\sqrt{2}$
Therefore, $\left(b-a\right)^3=16\sqrt{2}$.
Thus, the area, $S=\frac{8\sqrt{2}}{3}$.
And this answer was able to be posted quickly due to my MathJax friend, Desmos.
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