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If f is the function whose graph is shown in the figure below and g(x) = f(t) dt, find the values of g(0), g(1), g(2), g(3), g(4), and g(5). Then sketch a rough grapl 2. y= f(t) 1+ 1 4 Solution First, we notice that g(0) = f(t) dt = 0. From the figures below we see that g(1) is the area of a triangle. g(1) = f(t) dt = 1· 2) = To find g(2), we add to g(1) the area of a rectangle. g(2) = f(t) dt f(t) dt + f(t) dt = 1 + We estimate that the area under f from 2 to 3 is about 1.3, so

If f is the function whose graph is shown in the figure below and g(x) = f(t) dt, find the values of g(0), g(1), g(2), g(3), g(4), and g(5). Then sketch a rough grapl 2. y= f(t) 1+ 1 4 Solution First, we notice that g(0) = f(t) dt = 0. From the figures below we see that g(1) is the area of a triangle. g(1) = f(t) dt = 1· 2) = To find g(2), we add to g(1) the area of a rectangle. g(2) = f(t) dt f(t) dt + f(t) dt = 1 + We estimate that the area under f from 2 to 3 is about 1.3, so

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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We estimate that the area under f from 2 to 3 is about 1.3, so -3 g(3) = g(2) + J2 f(t) dt * 3 + 1.3 = 4.3. For t > 3, f(t) is negative and so we start subtracting areas. 4 g(4) = 9(3) 3 = 4.3 + (-1.3) '5 g(5) g(4) + | re) dt 14 = 3 + (-1.3)
Transcribed Image Text:We estimate that the area under f from 2 to 3 is about 1.3, so -3 g(3) = g(2) + J2 f(t) dt * 3 + 1.3 = 4.3. For t > 3, f(t) is negative and so we start subtracting areas. 4 g(4) = 9(3) 3 = 4.3 + (-1.3) '5 g(5) g(4) + | re) dt 14 = 3 + (-1.3)
If f is the function whose graph is shown in the figure below and g(x) = f(t) dt, find the values of g(0), g(1), g(2), g(3), g(4), and g(5). Then sketch a rough grapl 2. y= f(t) 1+ 1 4 Solution First, we notice that g(0) = f(t) dt = 0. From the figures below we see that g(1) is the area of a triangle. g(1) = f(t) dt = 1· 2) = To find g(2), we add to g(1) the area of a rectangle. g(2) = f(t) dt f(t) dt + f(t) dt = 1 + We estimate that the area under f from 2 to 3 is about 1.3, so
Transcribed Image Text:If f is the function whose graph is shown in the figure below and g(x) = f(t) dt, find the values of g(0), g(1), g(2), g(3), g(4), and g(5). Then sketch a rough grapl 2. y= f(t) 1+ 1 4 Solution First, we notice that g(0) = f(t) dt = 0. From the figures below we see that g(1) is the area of a triangle. g(1) = f(t) dt = 1· 2) = To find g(2), we add to g(1) the area of a rectangle. g(2) = f(t) dt f(t) dt + f(t) dt = 1 + We estimate that the area under f from 2 to 3 is about 1.3, so
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