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$-6-\int_{(-3)}^{(0)}\sqrt{9-x^{2}}dx$

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Find trigonometric ratios using reference angles

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Q.

-6-\int_{(-3)}^{(0)}\sqrt{9-x^{2}}dx

Understand the integral part: Understand the integral part of the expression. $\newline$ The integral $\int_{-3}^{0} \sqrt{9-x^2} \, dx$ represents the area under the curve of $y = \sqrt{9-x^2}$ from $x = -3$ to $x = 0$ . The equation $y = \sqrt{9-x^2}$ represents the upper semicircle of a circle with radius $3$ centered at the origin.
Calculate the area: Calculate the area under the curve. $\newline$ Since the curve is a semicircle with radius $3$ , the area of a full circle would be $\pi \times \text{radius}^2$ , which is $\pi \times 3^2 = 9\pi$ . The area under the curve from $x = -3$ to $x = 0$ is half of the full circle, so it is $\frac{9\pi}{2}$ .
Combine with rest of expression: Combine the integral result with the rest of the expression. $\newline$ Now we subtract the area we found from the constant term $-6$ to get the value of the entire expression: $-6 - (\frac{9\pi}{2})$ .
Simplify the expression: Simplify the expression. $\newline$ To simplify $-6 - \frac{9\pi}{2}$ , we can convert $-6$ to a fraction with the same denominator as $\frac{9\pi}{2}$ , which is $-\frac{12}{2}$ . Now the expression is $-\frac{12}{2}$ - $\frac{9\pi}{2}$ .
Combine the fractions: Combine the fractions. $\newline$ Combining the fractions $(-\frac{12}{2}) - (\frac{9\pi}{2})$ gives us $(-12 - 9\pi) / 2$ .

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