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Mth 174
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
Question - Problem 14 Ch.6.1
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Question: Section 6.1, Problem 5 5th edition Problem 14 4th edition Problem 5 [[6.1.5 (previously 6.1 #12)]]
f '(x) =1 for x on the interval (0,2), -1 on (2,3), 2 on (3,4), -2 on (4,6), 1 on (6,7)
f(3) = 0
What was your value for the integral of f '?
Your solution:
Confidence Assessment:
Given Solution:
the change in f from x=0 to x=2 is 2 (area beneath line segment from x=0 to x=1), then from x=2 to x=3 is -1, then from x=3 to x=4 is +2, then from x=4 to x=6 is -4, then from x=6 to x=7 is +1.
If f(3) = 0 then f(4) = 0 + 2 = 2, f(6) = 2 - 4 = -2 and f(7) = f(6) + 1 = -1.
Working back from x=3, f(2) = 0 - (-1) = 1 and f(0) = 1 - 2 = -1.
The integral is the sum of the changes in f ' which is 2 - 1 + 2 - 4 + 1 = 0.
Alternatively since f(0) = -1 and f(7) = -1 the integral is the difference f(7) - f(0) = -1 - (-1) = 0.
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I understand how to graph the antiderivative of this graph. I understand this statement: the change in f from x=0 to x=2 is 2 (area beneath line segment from x=0 to x=1), then from x=2 to x=3 is -1, then from x=3 to x=4 is +2, then from x=4 to x=6 is -4, then from x=6 to x=7 is +1.
I just don't understand why f(4) = 0 + 2 = 2, f(6) = 2 - 4 = -2 and f(7) = f(6) + 1 = -1. Why is f(4)associated with the area of 2 and f(6) with the area of - 4? Also, how is f(7) = f(6) + 1?
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This is a good question.
The antiderivative function f ' (x) can give you only the change in the value of the function f(x).
It is given that f(3) = 0.
The change in f(x) from x = 3 to x = 4 is 2, as you understand.
If f(3) = 0, then it follows that f(4) = 2.
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