Question

1. \int_{-3}^{3} \sqrt{9 - x^2} \\, dx \quad 16. \int_{\dots}^{0} \sqrt{\dots}

Explanation:

Step1: Recognize the integral as a circle equation

The integrand $\sqrt{9 - x^2}$ can be rewritten by squaring both sides (considering the domain where it's real, i.e., $9 - x^2\geq0$ or $- 3\leq x\leq3$) as $y=\sqrt{9 - x^2}$, then squaring gives $y^{2}=9 - x^{2}$, or $x^{2}+y^{2}=9$. This is the equation of a circle with radius $r = 3$ and centered at the origin $(0,0)$. And since $y=\sqrt{9 - x^2}\geq0$, the graph of the integrand is the upper - half of the circle $x^{2}+y^{2}=9$.

Step2: Interpret the definite integral as an area

The definite integral $\int_{a}^{b}f(x)dx$ represents the net area between the curve $y = f(x)$, the $x$ - axis, and the lines $x = a$ and $x = b$. In this case, $f(x)=\sqrt{9 - x^2}$, $a=-3$, and $b = 3$. Since $y=\sqrt{9 - x^2}\geq0$ for $-3\leq x\leq3$, the integral $\int_{-3}^{3}\sqrt{9 - x^2}dx$ represents the area under the curve $y=\sqrt{9 - x^2}$ from $x=-3$ to $x = 3$.

Step3: Calculate the area of the upper - half circle

The area of a full circle is given by the formula $A=\pi r^{2}$. For a circle with radius $r = 3$, the area of the full circle is $A=\pi\times(3)^{2}=9\pi$. The upper - half of the circle has an area that is half of the area of the full circle. So the area of the upper - half circle is $\frac{1}{2}\times9\pi=\frac{9\pi}{2}$.

Answer:

$\frac{9\pi}{2}$