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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.

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.

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.

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 $x$.