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Sagot :
The limit from 1 to 2 of the given antiderivative is; -0.19865
What is the Limit of the Integral?
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;
[tex]\int\limits^2_1 {(sin\frac{1}{x^{2} + 1} )} \, dx = {(sin\frac{1}{2^{2} + 1} )} - {(sin\frac{1}{1^{2} + 1} )}[/tex]
⇒ (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; https://brainly.com/question/10268976
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