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
So using a(2)=0 we can first solve for k by substituting t for 2
0 = (2-k)(2-3)(2-6)(2+3)
0 = (2-k)(-1)(-4)(5)
0 = (2-k)20
0 = 40 - 20k
-40 = -20k
k = 2
The next step would be to find all the 0s of a.
0 = (t-2)(t-3)(t-6)(t+3)
T = 2,3,6,-3
Then we find the product
2x3x6x-3 = -108
Since the problem asks for the absolute value, the answer is positive 108
Answer:
wears
Explanation:
because verbs are the ones which use the third person singular like she goes,he wears,he likes