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Answers for “如何设置定积分以求曲线 y = x^{3/2} 在区间 [0, 32/9] 上的弧长?”

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To set up the definite integral for the arc length of the curve y=x3/2y = x^{3/2} over the interval [0,32/9][0, 32/9], you first apply the power rule to find the derivative of the function, which is f′(x)=32x1/2f'(x) = \frac{3}{2}x^{1/2}. Next, you square this derivative to get (f′(x))2=94x(f'(x))^2 = \frac{9}{4}x.

Conditions: The curve is defined by the function f(x)=x3/2f(x) = x^{3/2}.; The interval of integration is [0,32/9][0, 32/9].; The arc length formula ∫ab1+(f′(x))2dx\int_{a}^{b} \sqrt{1 + (f'(x))^2} dx is applicable.

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For the arc length formula ∫ab1+(f′(x))2dx\int_{a}^{b} \sqrt{1 + (f'(x))^2} dx to be valid, the function f(x)f(x) must be continuous on the closed interval [a,b][a, b], and its derivative f′(x)f'(x) must also be continuous on [a,b][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)y = f(x).; The interval of integration is [a,b][a, b].; f(x)f(x) is continuous on [a,b][a, b].; f′(x)f'(x) is continuous on [a,b][a, b].