Asap For t > 3, f(t) is negative and so we start subtracting areas:[ƒ (1) dt ≈ 4.3 + (−1.3)g (5) = g(4) +· + f² f(t) dt 3+ (-1.3)g (4) = g(3) +We use these values to sketch the graph of g below. Notice that, because f(t) is positive for t < 3,we keep adding area for t <3 and so g is increasing up to x = 3 where it attains a maximumvalue. For x > 3, g is decreasing because f(t) is negative.3-A2-2341-14 5 y=f(t)Video Example)EXAMPLE 1 Let f be the function whose graph is shown in the figure to the left. Consider thefunction g(x) defined below.g(x) == f* 1 (1) diFind 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 thatg (0) = = [₁ 1 (1) dt = 0From the figure below we see that g(1) is the area of the triangle:= ['₁ f (t) dt =g (1) =g (2) == 1 +FI0-1=g(1) 101g (3)= g (2) +-1,22To find g(2), we add to g(1) the area of the rectangle:- [ f (t) dt = -6₁g(4) 3g(2)=30f (t) dt +fsf (t) dtWe estimate that the area under f from 2 to 3 is about 1.3, so+ f₂50For t > 3, f(t) is negative and so we start subtracting areas:f(t) dt 3+1.3 = 4.3g (4) = g(3) + - [* f (t) dt ≈ 4.3 + (−1.3)2g (5) = g(4) + ff (t) dt ≈ 3+ (−1.3)g(3) 4.3g(5) 1.7

Answered: Image /qna-images/answer/4760e591-f42c-4935-b6d4-2980561a4b5b.jpg

y=f(t) Video Example) EXAMPLE 1 Let f be the function whose graph is shown in the figure to the left. Consider the function g(x) defined below. g(x) = = f* ƒ (1) dt 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 = 6° ₁ From the figure below we see that g(1) is the area of the triangle: g(1) = ) = ['₁'ƒ ( * f (t) dt = 1; 2 = g (0) = g (2) = = 1+ 0 f(t)dt = 0 g(1) 1 0 = 1 To find g(2), we add to g(1) the area of the rectangle: • f ² ƒ (1) dt = [² ƒ (1) dt + f(t)dt Ss g(3) = g(2) + 2 g(4) 3 g (4) = g(3) + g(2) = 3 0 f(t) dt We estimate that the area under f from 2 to 3 is about 1.3, so + [ f(t) dt ≈ 3+1.3 = 4.3 For t > 3, f(t) is negative and so we start subtracting areas: [s f(t) dt 4.3+(-1.3) g(3) 4.3 g(5) 1.7 g (5) = g(4) + ff (t) dt ≈ 3+ (−1.3)