We are asked to evaluate the definite integral `int_0^2 (x^2-2x+3)dx`

Since `x^2-2x+3>0` for all x, this is also the area between the curve and the x-axis.

The simplest way is to apply the Fundamental Rule of Calculus: `int_(a)^(b)f(x)dx=F(b)-F(a)` where F is the antiderivative of f.

Since integrals are essentially sums, the integral of a sum is the sum of the integrals so we can integrate term by term: `int_0^2x^2dx + int_0^2 -2xdx + int_0^2 3dx`

`int_0^2 x^2dx=x^3/3 |_0^2=8/3-0=8/3`

The antiderivative of `x^2` is `x^3/3` ; if we differentiate `x^3/3` we get `x^2` so `F(x)=x^3/3,F(2)=8/3,F(0)=0`

`int_0^2 -2x=-2 int_0^2 xdx` (We can pull out constant multipliers.)

If `f(x)=x " then " F(x)=x^2/2` so `-2int_0^2 xdx=-2(x^2/2 |_0^2)=-2(2-0)=-4`

`int_0^23dx=3x|_0^2=6-0=6`

Thus `int_0^2 (x^2-2x+3)dx=(x^3/3-x^2+3x)|_0^2=8/3-4+6=14/3`

Another option is to use the Riemann sum definition of the definite integral:

`int_a^b f(x)dx=lim_(n->oo)sum_(i=1)^n f(c_(i))Deltax_(i)` where `Delta` is a partition of the interval and `c_i` is a value in the `i^(th)` subinterval. Using a regular partition of width `2/n` and choosing `c_(i)` to be the right endpoint of each subinterval we get:

`int_0^2 (x^2-2x+3)dx=lim_(n->oo)sum_(i=1)^n (((2i)/n)^2-2((2i)/n)+3)(2/n)`

Using properties of sums we can manipulate:

`=lim_(n->oo)(2/n^3)sum_(i=1)^n 4i^2-4 i n+3n^2`

`=lim_(n->oo)(2/n^3)(2/3(2n^3+3n^2+n)-2n^3-2n^2+3n^3)`

`=lim_(n->oo)[8/3+6/n+2/n^2-4-4/n+6]=14/3`

This can be interpreted as the area between the parabola defined by the function and the x-axis over the interval from 0 to 2. It can also be interpreted as an accumulation function.

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