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valentinak56 [21]
3 years ago
11

Find the average value of the function f(t)=cos13(5t)sin(5t) f(t)=cos13⁡(5t)sin⁡(5t) on the interval [4,10].

Mathematics
1 answer:
Vilka [71]3 years ago
6 0
The average value is given by

\displaystyle\frac1{10-4}\int_4^{10}\cos^{13}5t\sin5t\,\mathrm dt

Setting y=\cos5t, you get \mathrm dy=-5\sin5t\,\mathrm dt, and the integral becomes

\displaystyle-\frac1{30}\int_{\cos20}^{\cos50}y^{13}\,\mathrm dy
=-\dfrac1{30}\dfrac{y^{14}}{14}\bigg|_{y=\cos20}^{y=\cos50}
=-\dfrac{\cos^{14}50-\cos^{14}20}{420}\approx-0.00145
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Find an equation for the nth term of the arithmetic sequence.
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Answer:

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Step-by-step explanation:

* Lets revise the arithmetic sequence

- There is a constant difference between each two consecutive numbers

- Ex:

# 2  ,  5  ,  8  ,  11  ,  ……………………….

# 5  ,  10  ,  15  ,  20  ,  …………………………

# 12  ,  10  ,  8  ,  6  ,  ……………………………

* General term (nth term) of an Arithmetic sequence:

- U1 = a  ,  U2  = a + d  ,  U3  = a + 2d  ,  U4 = a + 3d  ,  U5 = a + 4d

- Un = a + (n – 1)d, where a is the first term , d is the difference

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, n is the position of the term

* Lets solve the problem

∵ an = a + (n - 1)d

∴ a14 = a + (14 - 1)d

∴ a14 = a + 13d

∵ a14 = -33

∴ a + 13d = -33 ⇒ (1)

- Similar we can find another equation from a15

∵ a15 = a + (15 - 1)d

∴ a15 = a + 14d

∵ a15 = 9

∴ a + 14d = 9 ⇒ (2)

- We will solve equations (1) and (2) to find a and d

* Lets subtract equation (2) from equation (1)

∴ (a - a) + (13 - 14)d = (-33 - 9)

∴ -d = -42 ⇒ × both sides by -1

∴ d = 42

- Substitute this value of d in equation (1) or (2)

∵ a + 13d = -33

∵ d = 42

∴ a + 13(42) = -33

∴ a + 546 = -33 ⇒ subtract 546 from both sides

∴ a = -579

* Now lets write the equation of the nth term

∵ an = a + (n - 1)d

∵ a = -579 and d = 42

∴ an = -579 + (n - 1) 42 ⇒ open the bracket

∴ an = -579 + 42n - 42

∴ an = -621 + 42n

* The equation of the nth term is an = -621 + 42n

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