Discussion
Total variation of the function :
\[ f(x) = \frac{x^{3}}{3} - x, \ x \in [2, 3] \]
is :
Answer: A
The total variation is 16/3.
For a differentiable function, the total variation on \([a, b]\) is \(\int_{a}^{b} |f'(x)|\, dx\). Here \(f'(x) = x^{2} - 1\), which is positive on [2, 3], so f is increasing and the variation is simply \(f(3) - f(2)\):
\[ \begin{aligned} f(3) &= 9 - 3 = 6 \\ f(2) &= \tfrac{8}{3} - 2 = \tfrac{2}{3} \\ V &= 6 - \tfrac{2}{3} = \tfrac{16}{3} \end{aligned} \]
0 would hold only for a constant function. 17/3 and 13/3 come from slips in \(f(2)\), for example taking it as \(1/3\) or \(5/3\).
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