We are asked to find the area between the x-axis and `f(x)=3x^2-6x-9` on the interval [-5,4]. By agreement, the area above the x-axis is considered positive, while beneath the x-axis is negative.
The most straightforward way to compute the area is to evaluate the definite integral `int_(-5)^4 (3x^2-6x-9)dx`
Using the Fundamental Theorem of Calculus, we evaluate the antiderivative of f(x) at the endpoints and find the difference of their values or compute F(4)-F(-5) where F is the antiderivative of f.
Since f is a polynomial we can integrate term by term to get
`int_(-5)^4 (3x^2-6x-9)dx=x^3-3x^2-9x |_(-5)^4`
Since a definite integral of a function of x with x restricted to the real numbers can be defined as a Riemann sum.
`int_(-5)^4 (3x^2-6x-9)dx=lim_(n->oo)sum_(i=1)^n f(c_(i)) Deltax_(i)` where `c_(i)` is a vlue in the `i^(th)` subinterval and `Delta` is a partition of the `x_(i)'"s"` .
If we use a regular partition of `Delta x=9/n` and choose `c_(i)` to be the right endpoint of each subinterval we get
`int_(-5)^4 (3x^2-6x-9)dx=lim_(n->oo)sum_(i=1)^n ( 3(-5+(9i)/n)^2-6(-5+(9i)/n)-9)(9/n)` `=lim_(n->oo)sum_(i=1)^n (2187i^2)/n^3-(2916i)/n^2+864/n`
(Here we use the properties of summations including the special sums of 1,i, and i^2 and the fact that if n is in the denominator, the limit goes to zero.)
We could break the area into 3 regions: [-5,-1],[-1,3],[3,4] if we wanted to account for which regions were above the x-axis and which were below.