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 +  =  .

Let g(x) = xf(t) dt0, where f is the function whose graph is shown.(a) Evaluate g(0), g(2), g(4), g(6), and g(12).g(0) = g(2) = g(4) = g(6) = g(12) =

If F(x) = xf(t) dt2, where f is the function whose graph is given, which of the following values is largest?F(0)F(1)    F(2)F(3)F(4)

The graph of f is shown. Evaluate each integral by interpreting it in terms of areas.(a)    10f(x) dx0 (b)    25f(x) dx0 (c)    35f(x) dx25 (d)    45f(x) dx0

The graph of a function f is shown below.Find f0.

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) =