alleryStart learning →Why does no arbitrary constant appear in the final value of a definite integral?
AIt cancels in g(b) - g(a)
BIt is absorbed in the limits
CIt is always taken as zero
DIt is integrated away
Answer & Solution
Correct answer: A. It cancels in g(b) - g(a)
The same constant c appears in both g(b) + c and g(a) + c, so it cancels on subtraction.
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