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Answers for “如何积分 $\sqrt{x}$?”

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The antiderivative of x\sqrt{x} is 23x3/2\frac{2}{3}x^{3/2}. This is found by rewriting x\sqrt{x} as x1/2x^{1/2} and applying the power rule for integration, which states that ∫xndx=xn+1n+1\int x^n dx = \frac{x^{n+1}}{n+1} for n≠−1n \neq -1.

Conditions: The function is x\sqrt{x}.; The power rule for integration is applied.; The exponent n=1/2n = 1/2.

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The exact value of the definite integral ∫−339−x2 dx\int_{-3}^{3} \sqrt{9-x^2}\,dx is 9π2\frac{9\pi}{2}. This is derived by interpreting the integral as the geometric area of the upper semicircle of a circle with radius 3.

Conditions: Interpret the integral as area under the graph on [−3,3][-3,3].; Recognize the graph as the upper semicircle of radius 3.; Exact equality for the displayed definite integral.

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The final value is 23−4232\sqrt{3} - \frac{4\sqrt{2}}{3}. This is obtained by evaluating the antiderivative 23x3/2\frac{2}{3}x^{3/2} at the converted bounds x=3x=3 and x=2x=2 and subtracting the results.

Conditions: The integral is ∫0ln⁡2eu1+eudu\int_0^{\ln 2} e^u \sqrt{1+e^u} du.; The substitution x=1+eux = 1 + e^u is used.; The bounds are converted to 2 and 3.; The antiderivative is 23x3/2\frac{2}{3}x^{3/2}.

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The real domain of the function y=9−x2y=\sqrt{9-x^2} is the closed interval [−3,3][-3, 3]. This restriction exists because, over the real numbers, the expression inside a square root (the radicand) must be nonnegative for the principal square root to be real-valued.

Conditions: Real-valued interpretation of the square root.; Using the principal square root convention.; The radicand is 9−x29 - x^2.

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The final numerical value of the area moment of inertia IyI_y for the specific shaded region about the centroidal y-axis is approximately 0.762 m40.762 \text{ m}^4. This value is obtained by evaluating the definite integral ∫02x2(2−2x1/2)dx\int_0^2 x^2 (2 - \sqrt{2} x^{1/2}) dx, which simplifies to 243−2(27)23.5\frac{2^4}{3} - \sqrt{2} \left(\frac{2}{7}\right) 2^{3.5}.

Conditions: The region is bounded by y=2y = 2 (top) and y=2x1/2y = \sqrt{2} x^{1/2} (bottom) from x=0x=0 to x=2x=2.; The axis is the centroidal y-axis.; Units are in meters.

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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].

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The curly d symbol (∂\partial) represents a partial derivative. It means taking the derivative of a multivariable function with respect to one specific variable at a time, while treating all other variables as constants.

Conditions: The function has more than one independent variable.; All variables except the differentiation variable are treated as constants during the partial derivative.