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Citrus2011 [14]
2 years ago
6

If sin(1/x^2+1) is an antiderivative for f(x) then.

SAT
1 answer:
STatiana [176]2 years ago
8 0

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;

\int\limits^2_1 {(sin\frac{1}{x^{2} + 1} )} \, dx = {(sin\frac{1}{2^{2} + 1} )}  - {(sin\frac{1}{1^{2} + 1} )}

⇒ (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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