Question

use the formulas for areas of triangles and circles to evaluate the integrals of the functions graphed in the figure. enter the sum of all shaded areas in terms of π. provide your answer below.

Explanation:

Step1: Analyze the shapes

The shaded region consists of a triangle and two semi - circles.

Step2: Calculate the area of the triangle

The triangle has a base \(b = 2\) and height \(h=3\). The area of a triangle \(A_{t}=\frac{1}{2}bh\). So \(A_{t}=\frac{1}{2}\times2\times3 = 3\).

Step3: Analyze the semi - circles

The two semi - circles together form one full circle. For the circle formed by the two semi - circles, we need to find the radius.
For the function \(g(x)=\sqrt{- 48 + 16x-x^{2}}\), we complete the square: \(-48 + 16x-x^{2}=-(x^{2}-16x + 48)=-(x - 6)(x - 10)\). The standard form of a circle equation is \((x - a)^{2}+(y - b)^{2}=r^{2}\). Completing the square for \(-48 + 16x-x^{2}\), we have \(-(x^{2}-16x + 64-16)=16-(x - 8)^{2}\). So the radius of the circle formed by the two semi - circles \(r = 4\).
The area of a circle \(A_{c}=\pi r^{2}\), with \(r = 4\), \(A_{c}=\pi\times4^{2}=16\pi\).
For the function \(h(x)=\sqrt{-192 + 28x-x^{2}}\), we complete the square: \(-192+28x - x^{2}=-(x^{2}-28x + 192)=-(x - 14)(x - 14)\). Completing the square: \(-(x^{2}-28x+196 - 4)=4-(x - 14)^{2}\), and the radius of this semi - circle \(r_{1}=2\). The area of this semi - circle \(A_{s1}=\frac{1}{2}\pi r_{1}^{2}=\frac{1}{2}\pi\times2^{2}=2\pi\).
The total area of the circles is \(A_{circles}=16\pi+2\pi=18\pi\).

Step4: Calculate the total shaded area

The total shaded area \(A = 3+18\pi\).

Answer:

\(3 + 18\pi\)