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To set up the definite integral for the arc length of the curve $y = x^{3/2}$ over the interval $[0, 32/9]$, you first apply the power rule to find the derivative of the function, which is $f'(x) = \frac{3}{2}x^{1/2}$. Next, you square this derivative to get $(f'(x))^2 = \frac{9}{4}x$.

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.

To set up the definite integral for the arc length of the curve $y = x^{3/2}$ over the interval $[0, 32/9]$, you first apply the power rule to find the derivative of the function, which is $f'(x) = \frac{3}{2}x^{1/2}$. Next, you square this derivative to get $(f'(x))^2 = \frac{9}{4}x$.

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.

For the arc length formula $\int_{a}^{b} \sqrt{1 + (f'(x))^2} dx$ to be valid, the function $f(x)$ must be continuous on the closed interval $[a, b]$, and its derivative $f'(x)$ must also be continuous on $[a, b]$. These conditions ensure that the curve is smooth enough for the integral to accurately represent its length.

Conditions: The curve is defined by a function $y = f(x)$.; The interval of integration is $[a, b]$.; $f(x)$ is continuous on $[a, b]$.; $f'(x)$ is continuous on $[a, b]$.

For the arc length formula $\int_{a}^{b} \sqrt{1 + (f'(x))^2} dx$ to be valid, the function $f(x)$ must be continuous on the closed interval $[a, b]$, and its derivative $f'(x)$ must also be continuous on $[a, b]$. These conditions ensure that the curve is smooth enough for the integral to accurately represent its length.

Conditions: The curve is defined by a function $y = f(x)$.; The interval of integration is $[a, b]$.; $f(x)$ is continuous on $[a, b]$.; $f'(x)$ is continuous on $[a, b]$.

The substitution $u = 1 + \frac{9}{4}x$ is chosen because it is exactly the expression under the radical in the arc-length integral $\int_0^{32/9}\sqrt{1+\frac{9}{4}x}\,dx$. By setting $u$ to this inner expression, the complicated integrand simplifies to $\sqrt{u}$, which is much easier to integrate using the power rule.

Conditions: The integral to evaluate is $\int_0^{32/9}\sqrt{1+\frac{9}{4}x}\,dx$.; The integrand contains a composite expression under a square root.

The substitution $u = 1 + \frac{9}{4}x$ is chosen because it is exactly the expression under the radical in the arc-length integral $\int_0^{32/9}\sqrt{1+\frac{9}{4}x}\,dx$. By setting $u$ to this inner expression, the complicated integrand simplifies to $\sqrt{u}$, which is much easier to integrate using the power rule.

Conditions: The integral to evaluate is $\int_0^{32/9}\sqrt{1+\frac{9}{4}x}\,dx$.; The integrand contains a composite expression under a square root.