Sketch the area represented by g(x).g(x) = xt2 dt1A curve and shaded region are graphed on the t y coordinate plane. The curve enters the viewing window in the second quadrant, goes down and right getting less steep. The curve passes through the origin, at which point it turns and goes up and right getting more steep and exits the viewing window at the top of the first quadrant. The region is below the curve and above the t-axis between the values t = 1 and t = x. A curve and a shaded region are graphed on the t y coordinate plane. The curve enters the viewing window in the third quadrant, goes up and right getting less steep. The curve passes through the origin, at thich point it goes up and right getting more steep and exits the viewing window at the top of the first quadrant. The region is below the curve and above the t-axis between the values t = 0 and t = 1. A curve and shaded region are graphed on the t y coordinate plane. The curve enters the viewing window in the second quadrant, goes down and right getting less steep. The curve passes through the origin, at which point it turns and goes up and right getting more steep and exits the viewing window at the top of the first quadrant. The region is below the curve and above the t-axis between the values t = 0 and t = 1. A curve and a shaded region are graphed on the t y coordinate plane. The curve enters the viewing window in the third quadrant, goes up and right getting less steep. The curve passes through the origin, at thich point it goes up and right getting more steep and exits the viewing window at the top of the first quadrant. The region is below the curve and above the t-axis between the values t = 1 and t = x.Find g ′(x) in two of the following ways.(a)by using part one of the fundamental theorem of calculusg ′(x) = (b)by evaluating the integral using part two of the fundamental theorem of calculus and then differentiatingg ′(x) =

The graph of a function g is shown.The x y-coordinate plane is given. A function labeled y = g(x) with 4 parts is graphed.The first part is a curve, enters the window in the second quadrant, goes up and right becoming less steep, crosses the y-axis at approximately y = 2.5, and ends at the open point (2, 3).The second part is a curve begins again at the open point (2, 1), goes up and right becoming less steep, and ends at the open point (5, 2).The third part is the closed approximate point (5, 1.2).The fourth part is a curve, begins at the open point (5, 2) goes down and right becoming more steep, and exits the window in the first quadrant.Use it to state the values (if they exist) of the following:(a)  lim x → 2− g(x)(b)  lim x → 2+ g(x)(c)  lim x → 2 g(x)(d)  lim x → 5− g(x)(e)  lim x → 5+ g(x)(f)  lim x → 5 g(x)SolutionLooking at the graph we see that the values of g(x) approach as x approaches 2 from the left, but they approach as x approaches 2 from the right.Therefore (a) lim x → 2− g(x) =     and    (b) lim x → 2+ g(x) = .Since the left and right limits are different, we conclude that (c) the limit as x approaches 2 of g(x) does not exist.The graph also shows that (d) lim x → 5− g(x) =     and    (e) lim x → 5+ g(x) = .This time, the left and right limits are the same and so, by this theorem, we have (f) lim x → 5 g(x) = Despite this fact, notice that g(5) ≠ 2.

Graphs of f and g are shown.There are two curves on the x y coordinate plane.A curve labeled f enters the top of second quadrant, goes down and right, crosses the negative x-axis and reaches a minimum in third quadrant, then goes up and right and intersects the origin, and reaches a maximum in first quadrant the same distance away from the x and y axes as its minimum was away from them. It then goes down and right, intersects the positive x-axis the same distance away from the origin as it intersected the negative x-axis, and exits the bottom of fourth quadrant, the same distance away from the x and y axes as was its entry point.A curve labeled g is always above the curve labeled f. g enters the top of second quadrant, goes down and right, reaches a minimum in second quadrant, goes up and right, then reaches a local maximum as it crosses the positive y-axis, and goes down and right. Then it reaches a minimum in first quadrant the same distance away from the x and y axes as was its first minimum. It then goes back up and exits the top of first quadrant the same distance away from the x and y axes as was its entry point.Is f even, odd, or neither?evenodd    neitherExplain your reasoning.It is symmetric about the origin.It is symmetric with respect to the y-axis.    It is symmetric with respect to the x-axis.It is not symmetric about the origin or the y-axis.Is g even, odd, or neither?evenodd    neitherExplain your reasoning.It is symmetric about the origin.It is symmetric with respect to the y-axis.    It is symmetric with respect to the x-axis.It is not symmetric about the origin or the y-axis.

Video ExampleEXAMPLE 1 If f is the function whose graph is shown in the figure to the left and g(x) = xf(t) dt0, find the values of g(0), g(1), g(2), g(3), g(4), and g(5). Then sketch a rough graph of g.SOLUTION First, we notice that g(0) = 0f(t) dt0 = 0. From the figure below we see that g(1) is the area of the triangle:g(1) = 1f(t) dt0 = 12(1 · 2) = .To find g(2), we add to g(1) the area of the rectangle:g(2)  =  2f(t) dt0 =  1f(t) dt0 + 2f(t) dt1 =  1 +  =  .

What is the area between two curves?

Let S be the solid obtained by rotating the region shown in the figure below about the y-axis.The x y-coordinate plane is given. There is a curve and a shaded region on the graph.The curve y = 3x(x − 1)2 starts at the origin, goes up and right becoming less steep, changes direction at the approximate point (0.33, 0.444), goes down and right becoming more steep, passes through the approximate point (0.67, 0.222), goes down and right becoming less steep, and ends at x = 1 on the positive x-axis.The shaded region is above the x-axis and below the curve from x = 0 to x = 1.Explain why it is difficult to use the washer method to find the volume V of S.This answer has not been graded yet.Sketch the solid. What are the circumference c and height h of a typical cylindrical shell?c(x) = 2πr h(x) = 3x(x−1)2 Use the method of cylindrical shells to find the volume V of S.