I've got the two function: f(x) = -4x + x³ and g(x) = 5x
they meet each other at -3, 0 and 3, where the areas between -3 and 0 and 0 and 3 are the same.
wolfram alpha link to the curves
what i want to do now, is using a single integal: $\int_{-3}^{3}\left | f(x)-g(x) \right |dx$ to calculate this area.
to calculate this id calculate a function $h(x) = f(x) - g(x) = x³ - 9x $, create the primitive function of $h$, which is $H(x) = \frac{1}{4} x^4 - 4.5x^2$
now, to calculate the integral:
$$ \int_{-3}^{3}\left | h(x) \right | = |H(3)| -| H(-3)| $$
but, at this point has to be a mistake, because $H(3) = H(-3)$ my result would always be $0$.
i could not find a clear solution to this problem. well to get the correct result either: the minus has to be a plus (in the case you use absolute values). or just it has to be $|H(3)| - H(-3)$.
but both seem a bit counter intuitive, because of the way the it is defined in the newton-leibniz formula.
i hopy someone can clarify this for my. thanks for the reading.
Both functions are odd. Hence $|f(x) -g(x)|$ is even. So the area: $$A = \int_{-3}^3 |f(x)-g(x)|= 2 \int_{0}^3 |f(x)-g(x)|\,.$$ As $f\le g$ on $[0,3]$ (which follows from their intersection and continuity), $$A = 2 \int_{0}^3 g(x)-f(x) = 2[\frac{9}{2}x^2-\frac{x^4}{4}]_0^3\,.$$
Now to avoid the computation of unnecessary fractions, remark at $x=3$ that $9x^2= x^4$, so: $$A = 2\times 81\times(\frac{1}{2}-\frac{1}{4}) = \frac{81}{2}\,.$$
Your mistake might be that if $\int h = H$, you do not have, in general, $\int |h| = |H|$, for existence reasons, but here mostly because of the sign change. This is why it is useful to split the interval in regions of constant sign.