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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.

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.

To evaluate $\int_2^3 \sqrt{x} dx$, first find the antiderivative $\frac{2}{3}x^{3/2}$. Then, apply the Fundamental Theorem of Calculus by subtracting the value of the antiderivative at the lower bound from its value at the upper bound: $\left[\frac{2}{3}x^{3/2}\right]_2^3 = \frac{2}{3}(3)^{3/2} - \frac{2}{3}(2)^{3/2}$.

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 $\int_2^3 \sqrt{x} dx$, first find the antiderivative $\frac{2}{3}x^{3/2}$. Then, apply the Fundamental Theorem of Calculus by subtracting the value of the antiderivative at the lower bound from its value at the upper bound: $\left[\frac{2}{3}x^{3/2}\right]_2^3 = \frac{2}{3}(3)^{3/2} - \frac{2}{3}(2)^{3/2}$.

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 use the curve $y^2 = 2x$ with vertical strips, solve for $y$ by taking the square root, yielding $y = \pm\sqrt{2x}$. Since the region lies in the first quadrant where $y \ge 0$, select the positive root $y = \sqrt{2} x^{1/2}$ to define the strip height as a function of $x$.

Conditions: The boundary curve is given as $y^2 = 2x$.; The region is in the first quadrant ($x \ge 0, y \ge 0$).; Vertical strips are being used, requiring height as a function of $x$.

To use the curve $y^2 = 2x$ with vertical strips, solve for $y$ by taking the square root, yielding $y = \pm\sqrt{2x}$. Since the region lies in the first quadrant where $y \ge 0$, select the positive root $y = \sqrt{2} x^{1/2}$ to define the strip height as a function of $x$.

Conditions: The boundary curve is given as $y^2 = 2x$.; The region is in the first quadrant ($x \ge 0, y \ge 0$).; Vertical strips are being used, requiring height as a function of $x$.

To evaluate the definite integral, substitute $x = 1 + e^u$, which gives $dx = e^u du$. Convert the bounds from $u=0$ and $u=\ln 2$ to $x=2$ and $x=3$.

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, substitute $x = 1 + e^u$, which gives $dx = e^u du$. Convert the bounds from $u=0$ and $u=\ln 2$ to $x=2$ and $x=3$.

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 substitution $x = 1 + e^u$ simplifies the integrand because its derivative $dx = e^u du$ exactly matches the $e^u du$ term in the original integral. This allows the complex expression $e^u \sqrt{1+e^u} du$ to be rewritten as the simpler radical $\sqrt{x} dx$.

Conditions: The integrand contains $e^u \sqrt{1+e^u}$.; The substitution is $x = 1 + e^u$.; The differential is $dx = e^u du$.

The substitution $x = 1 + e^u$ simplifies the integrand because its derivative $dx = e^u du$ exactly matches the $e^u du$ term in the original integral. This allows the complex expression $e^u \sqrt{1+e^u} du$ to be rewritten as the simpler radical $\sqrt{x} dx$.

Conditions: The integrand contains $e^u \sqrt{1+e^u}$.; The substitution is $x = 1 + e^u$.; The differential is $dx = e^u du$.

The substitution works because the derivative of $1+x^4$ is $4x^3$, which matches the $x^3$ term present in the integrand $x^3\sqrt{1+x^4}$. This allows the integral to be transformed into a simpler power function form, enabling the calculation of the antiderivative required for the integration by parts step.

Conditions: The term to be integrated is $x^3\sqrt{1+x^4}$.; The substitution variable is $w = 1+x^4$.; The differential $dw$ is proportional to $x^3 dx$.

The substitution works because the derivative of $1+x^4$ is $4x^3$, which matches the $x^3$ term present in the integrand $x^3\sqrt{1+x^4}$. This allows the integral to be transformed into a simpler power function form, enabling the calculation of the antiderivative required for the integration by parts step.

Conditions: The term to be integrated is $x^3\sqrt{1+x^4}$.; The substitution variable is $w = 1+x^4$.; The differential $dw$ is proportional to $x^3 dx$.

To set up the integral for the area moment of inertia about the y-axis ($I_y$) using a vertical strip, express the differential area $dA$ as the product of the strip's height (upper boundary minus lower boundary) and its width $dx$. Substitute this into the definition $I_y = \int x^2 dA$ and determine the integration limits based on the horizontal extent of the region bounded by functions of $x$.

Conditions: The region is bounded by curves that can be expressed as functions of $x$.; A vertical differential strip is chosen, meaning its width is $dx$.; The integration variable is $x$.

To set up the integral for the area moment of inertia about the y-axis ($I_y$) using a vertical strip, express the differential area $dA$ as the product of the strip's height (upper boundary minus lower boundary) and its width $dx$. Substitute this into the definition $I_y = \int x^2 dA$ and determine the integration limits based on the horizontal extent of the region bounded by functions of $x$.

Conditions: The region is bounded by curves that can be expressed as functions of $x$.; A vertical differential strip is chosen, meaning its width is $dx$.; The integration variable is $x$.

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.

To differentiate an integral with a square-root upper limit, you evaluate the integrand at the upper limit and multiply by the derivative of that limit. For an upper limit $u=\sqrt{t}$, the derivative is $1/(2\sqrt{t})$.

Conditions: The integrand is continuous on the relevant interval.; The upper limit is a differentiable function of $t$, specifically $u=\sqrt{t}$.; $t > 0$.

To differentiate an integral with a square-root upper limit, you evaluate the integrand at the upper limit and multiply by the derivative of that limit. For an upper limit $u=\sqrt{t}$, the derivative is $1/(2\sqrt{t})$.

Conditions: The integrand is continuous on the relevant interval.; The upper limit is a differentiable function of $t$, specifically $u=\sqrt{t}$.; $t > 0$.

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 $x$.