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2 years ago

15

(Long division)

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1 answer:

pav-90 [236]2 years ago

7 0

Answer:

hope it's helpful

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2 years ago

A)Determine whether the sequence is divergent or convergent. If it is convergent, evaluate its limit. If it diverges to infinity

Colt1911 [192]

Answer:

a) The sequence converges to 0

b) The lenght of the curve is $\frac{1}{54}(217^{3/2}-37^{3/2})$

Step-by-step explanation:

Consider the sequence $a_n = \frac{-6n^6 + \sin^2(7n)}{n^7+11}$

a) We will prove it using the sandwich lemma. Note that for all n $-1\leq \sin^2(7n)\leq 1$ , then

$\frac{-6n^6 -1}{n^7+11}\leq\frac{-6n^6 + \sin^2(7n)}{n^7+11}\leq \frac{-6n^6 + 1}{n^7+11}$

Note that the expressions on the left and the right hand side have a greater degree on the denominator than the one on the numerator. Then, by takint the limit n goes to infinty on both sides, we have that

$0 \leq\frac{-6n^6 + \sin^2(7n)}{n^7+11} \leq 0$

So, the sequence converges to 0.

b) The function $f(x) = 4x^{3/2}+7$ the formula of curve lenght is given by

$s = \int_a^b \sqrt[]{1+(f'(x))^2}dx$

in this case, a=1, b=6

Note that $f'(x) =6x^{{1/2}$ . Then

$s=\int_1^6 \sqrt[]{1+36x}dx$ . Take u = 1+36x. Then du= 36dx (i.e du/36 = dx). If x = 1, then u = 37 and if x = 6 then u = 217. So,

$s=\frac{1}{36}\int_{37}^{217}\sqrt[]{u} du = \frac{2}{36\cdot 3} \left.u^{3/2}\right|_{37}^{217}=\frac{1}{54}(217^{3/2}-37^{3/2})$

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