Answer
a) $\int_{0}^{2}$ $f(x)dx = -\pi$
b) $\int_{2}^{6}$ $f(x)dx = 4$
c) $\int_{-4}^{2}$ $f(x)dx = -2\pi-1$
d) $\int_{-4}^{6}$ $f(x)dx = -2\pi+3$
e) $\int_{-4}^{6}$ $|f(x)|dx = 2\pi+5$
f) $\int_{-4}^{6}$ $[f(x)+2]dx = -2\pi+23$
Work Step by Step
a) $\int_{0}^{2}$ $f(x)dx$. For this problem, look at the graph and notice that there is a fourth of a circle from $x = 0$ to $x = 2$. Therefore, find the area of the fourth of a circle by using the formula $A = \pi r^2$ and dividing the result by $4$ because it is only a fourth of a circle. Since $r = 2$ the area will be $\frac{\pi (2)^2}{4} = \frac{\pi 4}{4} = \pi$. The result is negative because the area is under the $x-$axis. The answer is $-\pi$
b) $\int_{2}^{6}$ $f(x)dx$. For this problem, look at the graph and notice that there is a triangle with height $2$ and base $4$ from $x = 2$ to $x = 6$. Find the area of the triangle with the formula $A = \frac{bh}{2} = \frac{2\times4}{2} = 4$. The answer is positive since the area is above the x axis.
c) $\int_{-4}^{2}$ $f(x)dx$. For this problem, look at the graph and notice that there is a semicircle and a triangle from $x = -4$ to $x = 2$. Find the area of the semicircle whose radius is 2 using the formula $A = \frac{\pi r^2}{2} = \frac{\pi (2)^2}{2} = 2\pi$. Since the semicircle is below the $x-$axis, the value of the integral is $-2\pi$. Now add to this result the area of the triangle with base $2$ and height $1$. Use the formula $A = \frac{bh}{2} = \frac{2\times1}{2} = 1$. Again, the triangle is also below the x axis, so the area is $-1$. The answer is $-2\pi-1$
d) $\int_{-4}^{6}$ $f(x)dx$ For this problem, we can use the property of integral addition to seperate the integral as shown:
$\int_{-4}^{6}f(x)dx = \int_{-4}^{2} f(x)dx + \int_{2}^{6} f(x)dx$.
Since we already know that $\int_{-4}^{2} f(x)dx = -2\pi-1$ and $\int_{2}^{6} f(x)dx = 4$, simply add the two values. The answer is $-2\pi-1 + 4 = -2\pi + 3$
e) $\int_{-4}^{6}$ $|f(x)|dx$ For this problem, the absolute value indicates that all negative areas become positive. Again, we can separate this integral into:
$\int_{-4}^{6}|f(x)|dx = \int_{-4}^{2} |f(x)|dx + \int_{2}^{6} |f(x)|dx$.
Since $\int_{-4}^{2} f(x)dx = -2\pi-1$, $\int_{-4}^{2} |f(x)|dx = 2\pi+1$ because all of the areas in the domain are negative.
