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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.