Question

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

Answer 1

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;

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

⇒ (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