P1 /2 1. Let's begin with a specific example. Consider the integral sin2 x dx. Graph the integrand f(x) = sin2x o on the interval [0, x/2] and identify the region whose area is given by the integral. Draw the rectangle with opposite vertices at (0, 0) and (7/2, 1) and compute its area. Argue that the graph of f(x) = sin x divides this rectangle in half and that the value of the integral is 7/4.2. The result of Step 1 depends on the assumption that the graph of y = f (x) divides the rectangle in .VA question into two regions of equal area. The goal is to understand when this assumption can be made. Here is the key idea. A function f is said to Symmetric about a point (p, q) if whenever the point (p - x, q - y) is on the graph offthen the point (p + x, q +y) is also on the graph. Said differently, f is symmetric about a point (p, q) if the line through the points (p, q) and (p + x, q + y) on the graph of y =f(x) intersects the graph at the point (p - x, q - y). Show that a function symmetric about the point (p, q) satises/(p - x) +f(p + x) = Zp) for all .1: in the interval of interest. 3. Suppose thatf is a continuous function on the interval [(1, b] and is symmetric about the point ((a + (3)2'2, (a + by?) (the point whose x-coordinate is the midpoint of [a, b] and whose ycoordinate is the corresponding point on the graph off). Using Step 2, show that f satises x) +f(a + b - x) = 2(a + b)a'2) for all x in [a, b]