Why does the presenter factor out $2/3$ before finishing the arc-length integral?

Conditions

• The expression to evaluate is $\frac{4}{9}\left[\frac{2}{3}\cdot 9^{3/2}-\frac{2}{3}\cdot 1^{3/2}\right]$.
• Both bracketed terms contain the same factor $\frac{2}{3}$.

Reasoning, step by step

1. Identify the common factor $\frac{2}{3}$ in both terms inside the brackets.
2. Factor out $\frac{2}{3}$ to rewrite the expression as $\frac{4}{9}\cdot\frac{2}{3}\left[9^{3/2}-1^{3/2}\right]$.
3. Multiply the outside constants $\frac{4}{9}$ and $\frac{2}{3}$ to get $\frac{8}{27}$.
4. Evaluate the powers inside the bracket: $9^{3/2} = 27$ and $1^{3/2} = 1$.
5. Subtract the values inside the bracket: $27 - 1 = 26$.
6. Multiply the result by the outside constant: $\frac{8}{27} \cdot 26 = \frac{208}{27}$.

Example

The speaker explicitly states, "Actually let's just factor out the two thirds. That makes it easier." The board then shows the transition from $\frac{4}{9}\left[\frac{2}{3}\cdot 9^{3/2}-\frac{2}{3}\cdot 1^{3/2}\right]$ to $\frac{8}{27}\cdot 26$.

Common misconceptions

• Attempting to distribute the $\frac{4}{9}$ into the brackets before factoring out the $\frac{2}{3}$, which complicates the arithmetic.
• Forgetting to multiply the factored-out $\frac{2}{3}$ by the outside constant $\frac{4}{9}$.
• Incorrectly evaluating fractional powers like $9^{3/2}$.

Watch the explanation

Explore next

• Why does $9^{3/2}-1^{3/2}$ become 26 in this worked example?
• What is the final exact value obtained for the displayed arc-length integral?