How do you set up the definite integral to find the arc length of the curve $y = x^{3/2}$ over the interval [0, $32/9$]?

Conditions

• The curve is defined by the function $f(x) = x^{3/2}$.
• The interval of integration is $[0, 32/9]$.
• The arc length formula $\int_{a}^{b} \sqrt{1 + (f'(x))^2} dx$ is applicable.

Reasoning, step by step

1. Identify the given function $f(x) = x^{3/2}$ and the interval $[0, 32/9]$.
2. Apply the power rule for derivatives to find $f'(x) = \frac{3}{2}x^{1/2}$.
3. Square the derivative to obtain $(f'(x))^2 = \frac{9}{4}x$.
4. Substitute $(f'(x))^2$ and the limits $a=0$ and $b=32/9$ into the arc length formula.
5. Write the final definite integral as $\int_{0}^{32/9} \sqrt{1 + \frac{9}{4}x} dx$.

Example

The video demonstrates this setup by stating the function $y = x^{3/2}$, applying the power rule to get $f'(x) = \frac{3}{2}x^{1/2}$, squaring it to get $\frac{9}{4}x$, and substituting these into the formula to establish the integral $\int_{0}^{32/9} \sqrt{1 + \frac{9}{4}x} dx$.

Common misconceptions

• Forgetting to square the derivative before substituting it into the arc length formula.
• Incorrectly applying the power rule, such as failing to multiply by the original exponent or subtracting one from it.
• Using the wrong limits of integration for the specific interval requested.

Watch the explanation

Explore next

• How do you evaluate the definite integral $\int_{0}^{32/9} \sqrt{1 + \frac{9}{4}x} dx$ using u-substitution?
• Why is the substitution $u = 1 + \frac{9}{4}x$ chosen for this arc-length integral?