Solve the problem from the image---**Problem Description:**
2. The shaded region R is bounded by the graphs of the functions f and g, where f(x) = x^2 - 2x and g(x) = x + sin(πx), as shown in the figure.
**(Note: Your calculator should be in radian mode.)**
A. Find the area of R. Show the setup for your calculations.
B. Region R is the base of a solid. For this solid, at each x the cross section perpendicular to the x-axis is a rectangle with height x and base in region R. Find the volume of the solid. Show the setup for your calculations.
C. Write, but do not evaluate, an integral expression for the volume of the solid generated when the region R is rotated about the horizontal line y = -2.
D. It can be shown that g'(x) = 1 + π cos(πx). Find the value of x, for 0 < x < 1, at which the line tangent to the graph of f is parallel to the line tangent to the graph of g.
**Chart/Diagram Description:**
* **Type:** 2D Cartesian coordinate graph.
* **Coordinate Axes:** X-axis labeled 'x' and Y-axis labeled 'y'. The origin is labeled 'O'.
* **Scales:** X-axis has tick marks at 1, 2, 3. Y-axis has tick marks at -2, -1, 1, 2, 3, 4.
* **Curves:**
* A parabolic curve labeled 'y = f(x)'. It passes through (0,0) and (2,0) and has a minimum around (1, -1).
* A curve labeled 'y = g(x)'. It passes through (0,0) and goes upwards, then downwards.
* **Intersection Points:** The curves intersect at the origin (0,0) and at the point labeled (3, 3).
* **Shaded Region:** The region R is shaded between the curves y = g(x) (upper boundary) and y = f(x) (lower boundary) from x = 0 to x = 3. The label 'R' is placed within the shaded region.
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We have two functions: f of x equals x squared minus 2x, and g of x equals x plus sine of pi x. These functions intersect at x equals 0 and x equals 3. The shaded region R is bounded by g of x above and f of x below, from x equals 0 to x equals 3. We need to solve four parts involving this region.
For part A, we need to find the area of region R. The area is calculated as the integral from 0 to 3 of g of x minus f of x. This gives us the integral of negative x squared plus 3x plus sine of pi x. Evaluating this integral, we get 9 halves plus 2 over pi.
For part B, we have a solid with base R and rectangular cross-sections perpendicular to the x-axis. Each cross-section has base equal to g of x minus f of x, and height equal to x. The volume is the integral from 0 to 3 of x times the difference of the functions, which equals 27 fourths plus 3 over pi.
For part C, when rotating region R about the line y equals negative 2, we use the washer method with outer radius g of x plus 2 and inner radius f of x plus 2. For part D, we find where the tangent lines are parallel by setting the derivatives equal. This gives us 2x minus 3 equals pi cosine of pi x, with solution x approximately 0.673.
To summarize our complete solution: Part A gives us an area of 9 halves plus 2 over pi. Part B yields a volume of 27 fourths plus 3 over pi. Part C provides the integral setup for rotation about y equals negative 2. And Part D finds the parallel tangent point at x approximately 0.673. These solutions demonstrate key calculus techniques including integration, cross-sectional volume methods, washer method for rotation, and derivative analysis.