# Determine the area limited by curves: $f(x)=2x^3-3x^2+9x$ and $g(x)=x^3-2x^2-3x$

Determine the area limited by curves:

$$f(x)=2x^3-3x^2+9x \\ g(x)=x^3-2x^2-3x$$

The correctly answer is: 25, How can I find it?

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Could you clarify the question? We usually talk about area of regions, not of functions. Are you looking for the area of some region that is bounded by the graphs of those functions? – MJD May 30 '12 at 18:36
Sorry, the correctly question is: Determine the limited area by curves f(x)=2x^3−3x^2+9x and g(x)=x^3−2x^2−3x – Alfredo May 30 '12 at 18:40
@Alfredo please edit your question accordingly – Belgi May 30 '12 at 18:40
The correctly answer is: 25. But how can I find it? – Alfredo May 30 '12 at 18:44
You need more information. $f$ and $g$ do not enclose any bounded area. – copper.hat May 30 '12 at 18:46

Assuming that the limits of the interval are given $a<b \in\mathbb{R}$, the fundamental theorem of calculus says that the "area" below the curve $f(x)$, and between $a$ and $b$ equals to $F(b)-F(a)$ where $F'(x)=f(x)$, $\forall x\in[a,b]$ (in case that such $F$ exists). Gladly, our $f$ & $g$ are polynomials and very easy to find an anti-derivative to (=indefinite integral).
So, we can find the anti-derivatives for $f(x)$ and $g(x)$ and evaluate the difference at $a$, $b$. If we let $S$ to be the area below $f(x)$ and above $g(x)$ we need to calculate the area below $f(x)$ and above the $X$ axis minus the area below $g(x)$ above the $X$ axis: $$S = F(x)-G(x)|_{x=a}^{x=b}$$
Notice that $\int f(x)+g(x) =\int f(x)+ \int g(x)$ so, you can integrate every "monic" by itself. Now, let $a\in\mathbb{R}$, $(a\cdot x^n)' = a\cdot n \cdot x^{n-1} \Rightarrow \int a\cdot x^{n-1} = a\frac{x^n}{n}$. – Amihai Zivan May 30 '12 at 19:41