question 7 of 13 use the figure to find the value of the integrals. note: enter exact answers, or round to…

Explanation:

Step1: Recall integral - area relationship

The definite integral $\int_{a}^{b}f(x)dx$ is equal to the net - signed area between the curve $y = f(x)$ and the $x$ - axis from $x=a$ to $x = b$. Area above the $x$ - axis is positive and area below the $x$ - axis is negative.

Step2: Analyze $\int_{9}^{21}f(x)dx$

We can split the interval $[9,21]$ into sub - intervals. From the graph, we can calculate the areas of geometric shapes (rectangles, triangles) formed by the curve and the $x$ - axis. The area from $x = 9$ to $x=12$ (a triangle below the $x$ - axis with base $3$ and height $3$) has area $A_1=-\frac{1}{2}\times3\times3= - 4.5$. The area from $x = 12$ to $x = 18$ (a semi - circle above the $x$ - axis with radius $3$) has area $A_2=\frac{1}{2}\pi r^{2}=\frac{1}{2}\pi\times3^{2}=\frac{9\pi}{2}\approx14.137$. The area from $x = 18$ to $x = 21$ (a triangle above the $x$ - axis with base $3$ and height $3$) has area $A_3=\frac{1}{2}\times3\times3 = 4.5$. Then $\int_{9}^{21}f(x)dx=A_1 + A_2+A_3=-4.5+\frac{9\pi}{2}+4.5=\frac{9\pi}{2}\approx14.137$.

Step3: Analyze $\int_{6}^{21}f(x)dx$

We know that $\int_{6}^{9}f(x)dx$ (a triangle below the $x$ - axis with base $3$ and height $3$) has area $A_4=-\frac{1}{2}\times3\times3=-4.5$. We already found $\int_{9}^{21}f(x)dx=\frac{9\pi}{2}$. So $\int_{6}^{21}f(x)dx=\int_{6}^{9}f(x)dx+\int_{9}^{21}f(x)dx=-4.5+\frac{9\pi}{2}\approx - 4.5 + 14.137=9.637$.

Step4: Analyze $\int_{15}^{24}f(x)dx$

The area from $x = 15$ to $x = 18$ (part of the semi - circle above the $x$ - axis) and from $x = 18$ to $x = 21$ (a triangle above the $x$ - axis) and from $x = 21$ to $x = 24$ (a rectangle below the $x$ - axis with base $3$ and height $3$). The area from $x = 15$ to $x = 18$ (a quarter - circle above the $x$ - axis with radius $3$) has area $A_5=\frac{1}{4}\pi r^{2}=\frac{9\pi}{4}\approx7.069$. The area from $x = 18$ to $x = 21$ (a triangle above the $x$ - axis with base $3$ and height $3$) has area $A_6 = 4.5$. The area from $x = 21$ to $x = 24$ (a rectangle below the $x$ - axis with base $3$ and height $3$) has area $A_7=-3\times3=-9$. Then $\int_{15}^{24}f(x)dx=A_5 + A_6+A_7=\frac{9\pi}{4}+4.5-9=\frac{9\pi}{4}-4.5\approx7.069 - 4.5 = 2.569$.

Answer:

(b) $\frac{9\pi}{2}\approx14.137$ (c) $\frac{9\pi}{2}-4.5\approx9.637$ (d) $\frac{9\pi}{4}-4.5\approx2.569$