Recall that the integral of the area between the graph of two functions, in an interval [a,b] is:
[tex]\int ^b_a|f(x)-h(x)|dx\text{.}[/tex]
Now, if f(x) is an odd function, we can use the following property:
[tex]\int ^a_{-a}|f(x)|dx=2\int ^a_0|f(x)|dx\text{.}[/tex]
Now, notice that the function y=-12x³ is an odd function, therefore:
[tex]\int ^1_{-1}|y-0|dx=2\int ^1_0|-12x^3|dx=2\int ^1_012x^3dx\text{.}[/tex]
Applying the linearity of the integral we get:
[tex]24\int ^1_0x^3dx=24\frac{x^4}{4}|^1_0=24(\frac{1}{4}-0)=\frac{24}{4}=6.[/tex]
Answer: 6.
