Question

in exercises 30 - 32, evaluate the integral by interpreting it and using formulas from geometry.

1. $int_{0}^{2}sqrt{4 - x^{2}}dx$
2. $int_{-4}^{8}|x|dx$
3. $int_{-8}^{8}2sqrt{64 - x^{2}}dx$

Explanation:

Step1: Analyze integral 30

The integral $\int_{0}^{2}\sqrt{4 - x^{2}}dx$ represents the area of a quarter - circle. The equation $y = \sqrt{4 - x^{2}}$ can be rewritten as $x^{2}+y^{2}=4$ ($y\geq0$), which is a semi - circle centered at the origin with radius $r = 2$. The integral from $x = 0$ to $x = 2$ gives a quarter - circle. The area of a circle is $A=\pi r^{2}$, so for a quarter - circle with $r = 2$, the area is $\frac{1}{4}\pi r^{2}$.
\[A_{30}=\frac{1}{4}\pi(2)^{2}=\pi\]

Step2: Analyze integral 31

The function $y = |x|$ is a V - shaped function. The integral $\int_{-4}^{8}|x|dx=\int_{-4}^{0}-x dx+\int_{0}^{8}x dx$. The integral $\int_{-4}^{0}-x dx$ represents the area of a right - triangle with base and height equal to 4, and its area is $\frac{1}{2}(4)(4) = 8$. The integral $\int_{0}^{8}x dx$ represents the area of a right - triangle with base and height equal to 8, and its area is $\frac{1}{2}(8)(8)=32$. So $\int_{-4}^{8}|x|dx=8 + 32=40$.

Step3: Analyze integral 32

The integral $\int_{-8}^{8}2\sqrt{64 - x^{2}}dx$. The equation $y=\sqrt{64 - x^{2}}$ can be rewritten as $x^{2}+y^{2}=64$ ($y\geq0$), which is a semi - circle centered at the origin with radius $r = 8$. The integral $2\int_{-8}^{8}\sqrt{64 - x^{2}}dx$ represents the area of a full circle with radius $r = 8$ (since the factor of 2 accounts for the two semi - circles above and below the x - axis). The area of a circle is $A=\pi r^{2}$, so $A_{32}=2\times\frac{1}{2}\pi(8)^{2}=64\pi$.

Answer:

1. $\pi$
2. $40$
3. $64\pi$