If sin(1/x^2+1) is an antiderivative for f(x) then.
1 answer:
The limit from 1 to 2 of the given antiderivative is; -0.19865
<h3>What is the Limit of the Integral?</h3>
We are given the antiderivative of f(x) as sin(1/(x² + 1)). Thus, to find the limit from 1 to 2, we will solve as;
⇒ (sin ¹/₅) - (sin ¹/₂)
⇒ 0.19866 - 0.47942
⇒ -0.19865
Complete Question is;
If sin(1/(x² + 1)) is an anti derivative for f(x), then what is the limit of f(x)dx from 1 to 2?
Read more about integral limits at; brainly.com/question/10268976
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