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?
 A: This problem has a nice solution without Calculus.

Let $P$ and $Q$ the intersections between the given parabola and the given line. Let $T$ be the intersection of the tangents at $P$ and $Q$. By Archimedes' lemma, the area of the wanted parabolic segment is just two thirds of the area of $PQT$. We have $T=(1;3)$ and $PQ=4$ by straightforward computations. The distance of $T$ from the $PQ$ line equals $2\sqrt{2}$, hence $[PQT]=\frac{1}{2}\cdot 4\cdot 2\sqrt{2}=4\sqrt{2}$ and the wanted area equals $\color{blue}{\frac{8}{3}\sqrt{2}}$.
Yet another interesting approach comes from noticing that Simpson's rule is an exact quadrature rule for quadratic polynomials, hence the given problem can be solved by just computing the coordinates of $P,Q$ and by evaluating $-x^2+3x-1$ at $x=1$, since that is the average of the abscissas of $P$ and $Q$.
A: 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.
A: at the first step you must compute the intersection Points of both functions
this leads to the equation
$$-x^2+2x+1=0$$
can you solve this?
for your Control we get
$$x_1=1-\sqrt{2}$$
$$x_2=1+\sqrt{2}$$
and then the searched area is given by $$A=\int_{x_1}^{x_2} -x^2+3x-1-(x-2)dx$$
A: This can be done using your suggested integration method. To find the intersections we set $x^{2}-3x+1=2-x$ so we have $x^{2}-2x-1=0$ so our limits of integration are the solutions to this quadratic namely $x=1\pm\sqrt{2}$.
Now to find the area we want the integral $\int_{1-\sqrt{2}}^{1+\sqrt{2}}-x^{2}+2x+1dx=[\frac{x^{3}}{3}+x^{2}+x]_{1-\sqrt{2}}^{1+\sqrt{2}}=\frac{8\sqrt{2}}{3}$
A: This is a simple integration problem where the required steps are:


*

*define your domain by:
a. finding the graphs intersection points.
b. understanding which graph is above the other (simple plot always helps).

*double-integrate "1" over your domain.


In your case, you will have $1-\sqrt{2}$ and $1+\sqrt{2}$ as your intersection points, and within this range the parabola lies above the line, so you have:
$$\int_{1-\sqrt{2}}^{1+\sqrt{2}}\int_{x-2}^{-x^2+3x-1} 1 dy dx$$ 
= $$ \int_{1-\sqrt{2}}^{1+\sqrt{2}} (-x^2+3x-1)-(x-2)  dx$$ = 
$$ (-x^3 / 3 + x^2 + x)\Bigr|_{1-\sqrt{2}}^{1+\sqrt{2}} $$
 $\approx$ 3.77
