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
I AM TOTAALLY BORED WITH SCHOOL
Answer:
Explanation:
the equation you use here is m=fa or mass=force x acceleration
you plug in the 13.5 N as the force, 6.5 m/s^2 as the acceleration, and do the math.
13.5 x 6.5
That would equal 87.75 which would be your answer.
The exercise is related to the Ratio Test for Convergence. The rules for this kind of test are given below.
<h3>What are the rules for Ratio Test for Convergence?</h3>
The rules are:
- If the limit is less than 1 when conducting the ratio test, your series is definitely convergent.
- The test is inconclusive if the limit is equal to 1.
- The series is divergent if the limit is greater than 1.
Using this knowledge, we are able to state that
- A is not conclusive.
- C's convergence is absolute.
- There is divergence with D and E.
- B and F appear to be employing the nth-term test. The nth-term involves determining the sequence's limit as it approaches infinity.
The nth-term test determines whether a series is divergent if the limit is bigger than 0, thus, both B and F are divergent series.
Please see the attached for the full question and the link below for more about Ratio Test for Convergence:
brainly.com/question/16618162
