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To evaluate $\int_2^3 \sqrt{x} dx$, find the antiderivative $\frac{2}{3}x^{3/2}$ and apply the Fundamental Theorem of Calculus. Subtract the value at the lower bound from the value at the upper bound to obtain the simplified result $2\sqrt{3} - \frac{4\sqrt{2}}{3}$.

Conditions: The integral is $\int_2^3 \sqrt{x} dx$.; The antiderivative is $\frac{2}{3}x^{3/2}$.; The bounds are 2 and 3.

To evaluate $\int_2^3 \sqrt{x} dx$, find the antiderivative $\frac{2}{3}x^{3/2}$ and apply the Fundamental Theorem of Calculus. Subtract the value at the lower bound from the value at the upper bound to obtain the simplified result $2\sqrt{3} - \frac{4\sqrt{2}}{3}$.

Conditions: The integral is $\int_2^3 \sqrt{x} dx$.; The antiderivative is $\frac{2}{3}x^{3/2}$.; The bounds are 2 and 3.

The definite integral can be evaluated by recognizing that the integrand $y = \sqrt{9-x^2}$ graphs as an upper semicircle of radius 3. Because the function is continuous and nonnegative on $[-3, 3]$, the integral equals the ordinary geometric area of this region.

Conditions: The integrand is recognized as the upper semicircle of $x^2+y^2=9$.; The function is continuous and nonnegative on the interval $[-3, 3]$.; Use the real geometric area formula for a circle.

The definite integral can be evaluated by recognizing that the integrand $y = \sqrt{9-x^2}$ graphs as an upper semicircle of radius 3. Because the function is continuous and nonnegative on $[-3, 3]$, the integral equals the ordinary geometric area of this region.

Conditions: The integrand is recognized as the upper semicircle of $x^2+y^2=9$.; The function is continuous and nonnegative on the interval $[-3, 3]$.; Use the real geometric area formula for a circle.

The equation $x^2 + y^2 = 9$ describes a full circle centered at the origin with radius 3, which includes both upper and lower branches. However, the original function is defined using the principal square root, $y = \sqrt{9-x^2}$.

Conditions: Working over the real numbers.; Using the principal (nonnegative) square root convention.; The underlying relation is the circle $x^2 + y^2 = 9$.

The equation $x^2 + y^2 = 9$ describes a full circle centered at the origin with radius 3, which includes both upper and lower branches. However, the original function is defined using the principal square root, $y = \sqrt{9-x^2}$.

Conditions: Working over the real numbers.; Using the principal (nonnegative) square root convention.; The underlying relation is the circle $x^2 + y^2 = 9$.

To evaluate the definite integral $\int_0^{\ln 2} e^u \sqrt{1+e^u} \, du$, use the substitution $x = 1 + e^u$, which yields $dx = e^u \, du$. Convert the limits of integration from $u=0$ and $u=\ln 2$ to $x=2$ and $x=3$, transforming the integral into $\int_2^3 \sqrt{x} \, dx$. Evaluating the antiderivative $\frac{2}{3}x^{3/2}$ at these new bounds gives the result $2\sqrt{3} - \frac{4\sqrt{2}}{3}$.

Conditions: The integral is definite with bounds $u=0$ to $u=\ln 2$.; The integrand is $e^u \sqrt{1+e^u}$.; Substitution $x = 1 + e^u$ is used.

To evaluate the definite integral $\int_0^{\ln 2} e^u \sqrt{1+e^u} \, du$, use the substitution $x = 1 + e^u$, which yields $dx = e^u \, du$. Convert the limits of integration from $u=0$ and $u=\ln 2$ to $x=2$ and $x=3$, transforming the integral into $\int_2^3 \sqrt{x} \, dx$. Evaluating the antiderivative $\frac{2}{3}x^{3/2}$ at these new bounds gives the result $2\sqrt{3} - \frac{4\sqrt{2}}{3}$.

Conditions: The integral is definite with bounds $u=0$ to $u=\ln 2$.; The integrand is $e^u \sqrt{1+e^u}$.; Substitution $x = 1 + e^u$ is used.

The antiderivative of $\sqrt{x}$ is $\frac{2}{3}x^{3/2}$, derived by rewriting the function as $x^{1/2}$ and applying the power rule for integration. This result is specifically used to evaluate definite integrals after transforming the bounds, where the constant of integration is omitted.

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

The antiderivative of $\sqrt{x}$ is $\frac{2}{3}x^{3/2}$, derived by rewriting the function as $x^{1/2}$ and applying the power rule for integration. This result is specifically used to evaluate definite integrals after transforming the bounds, where the constant of integration is omitted.

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

To evaluate the definite integral $\int_{-3}^{3} \sqrt{9-x^2}\,dx$ geometrically, recognize that the integrand $y = \sqrt{9-x^2}$ represents the upper semicircle of a circle centered at the origin with radius 3. Because the function is nonnegative and continuous on the interval $[-3, 3]$, the definite integral equals the ordinary geometric area of this shaded region.

Conditions: The integrand is $\sqrt{9-x^2}$ and the limits of integration are -3 and 3.; The square root denotes the principal (nonnegative) root, restricting the graph to $y \ge 0$.; The function is continuous and nonnegative on the closed interval $[-3, 3]$, ensuring the definite integral equals the ordinary area under the curve.

To evaluate the definite integral $\int_{-3}^{3} \sqrt{9-x^2}\,dx$ geometrically, recognize that the integrand $y = \sqrt{9-x^2}$ represents the upper semicircle of a circle centered at the origin with radius 3. Because the function is nonnegative and continuous on the interval $[-3, 3]$, the definite integral equals the ordinary geometric area of this shaded region.

Conditions: The integrand is $\sqrt{9-x^2}$ and the limits of integration are -3 and 3.; The square root denotes the principal (nonnegative) root, restricting the graph to $y \ge 0$.; The function is continuous and nonnegative on the closed interval $[-3, 3]$, ensuring the definite integral equals the ordinary area under the curve.

The definite integral $\int_0^{\ln 2} e^u \sqrt{1+e^u} du$ evaluates to the exact value $2\sqrt{3} - \frac{4\sqrt{2}}{3}$. This result is obtained by applying the substitution $x = 1 + e^u$, which transforms the bounds to 2 and 3, and then evaluating the antiderivative $\frac{2}{3}x^{3/2}$ at these new limits.

Conditions: The integral is $\int_0^{\ln 2} e^u \sqrt{1+e^u} du$.; The substitution $x = 1 + e^u$ is used.; The bounds are converted to 2 and 3.; The antiderivative is $\frac{2}{3}x^{3/2}$.

The definite integral $\int_0^{\ln 2} e^u \sqrt{1+e^u} du$ evaluates to the exact value $2\sqrt{3} - \frac{4\sqrt{2}}{3}$. This result is obtained by applying the substitution $x = 1 + e^u$, which transforms the bounds to 2 and 3, and then evaluating the antiderivative $\frac{2}{3}x^{3/2}$ at these new limits.

Conditions: The integral is $\int_0^{\ln 2} e^u \sqrt{1+e^u} du$.; The substitution $x = 1 + e^u$ is used.; The bounds are converted to 2 and 3.; The antiderivative is $\frac{2}{3}x^{3/2}$.

The real domain of the function $y=\sqrt{9-x^2}$ is the closed interval $[-3, 3]$. 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 $9 - x^2$.

The real domain of the function $y=\sqrt{9-x^2}$ is the closed interval $[-3, 3]$. 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 $9 - x^2$.

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

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

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

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

For a prime number $p$, the only positive divisors are 1 and $p$ itself. Any integer $a$ in the range from 1 to $p$ that is not equal to $p$ will not share the factor $p$.

Conditions: $p$ is a prime number.; $\varphi(p)$ is defined as the count of integers from 1 to $p$ that are coprime to $p$.

For a prime number $p$, the only positive divisors are 1 and $p$ itself. Any integer $a$ in the range from 1 to $p$ that is not equal to $p$ will not share the factor $p$.

Conditions: $p$ is a prime number.; $\varphi(p)$ is defined as the count of integers from 1 to $p$ that are coprime to $p$.

When using u-substitution in a definite integral, the original limits of integration in terms of $x$ must be converted into new limits in terms of $u$. This is done by substituting the original $x$-bounds into the substitution equation $u(x)$.

Conditions: The integral is a definite integral.; A substitution $u = u(x)$ is being used.; The original limits of integration are given in terms of $x$.

When using u-substitution in a definite integral, the original limits of integration in terms of $x$ must be converted into new limits in terms of $u$. This is done by substituting the original $x$-bounds into the substitution equation $u(x)$.

Conditions: The integral is a definite integral.; A substitution $u = u(x)$ is being used.; The original limits of integration are given in terms of $x$.

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.