Given the following integral I(n) = (1-2)"dx. We want to ultimately show that 22n+1 (n!)2 I(n) = (2n+1)! I 1. Use I(n) = f₁(1 − x²)" dr to evaluate: 1.1 I(1) 1.2 I(2) 2. Recall that I(n) = f₁(1 - x²)" dx. 2.1 Substitute x = sin(t) in I(n) and show that I(n) can be written as π/2 L -*/2 (cos(t))2n+1dt I(n) = 2.2 Now that we have I(n) = 1/2 (cos(t))2n+1dt. What is I(n - 1)? Leave your answer in integral form. /2 2.3 Next, integrate (cos(t))2n+1dt and show that it can be written as 2n I(n) = I(n - 1) 2n + 1 Hint: Integrate by parts. Split the exponent so that dv = cos(t)dt 3. Now that we have derived I(n) = 22(n-1), and since I(1) = 4/3 3.1 Confirm I(2)= 15 by using the new formula I(n) 16 2n 2n+1 (n - 1) 4. If we calculate I(3) using I(n) = 211(n - 1), we observe that 1(3) = -1 (2) I 3.3.2, and since I(2) = 16 or 2, then 642 - I(3) = 7 5 3 2 (This is known as the Wallis Product) Using I(n) =22I(n - 1) formula, calculate: 4.1 I(4) 4.2 I(5) (For 4.1 and 4.2, leave your answers in fractional format) 5. You should have observed that the general pattern using the fractional format is: I(n) = 2 2n 2n+1 8642 9 7 53 5.1 Use this fractional pattern to show that: I(n) = 22n+1 (n!)² (2n+1)! Recall that n! = n (n-1) (n-2)-4-3.2.1 Hint: Multiply the fractional pattern by 8 20100 8 2n and modify 2n the newly formed expression to its final form.
Given the following integral I(n) = (1-2)"dx. We want to ultimately show that 22n+1 (n!)2 I(n) = (2n+1)! I 1. Use I(n) = f₁(1 − x²)" dr to evaluate: 1.1 I(1) 1.2 I(2) 2. Recall that I(n) = f₁(1 - x²)" dx. 2.1 Substitute x = sin(t) in I(n) and show that I(n) can be written as π/2 L -*/2 (cos(t))2n+1dt I(n) = 2.2 Now that we have I(n) = 1/2 (cos(t))2n+1dt. What is I(n - 1)? Leave your answer in integral form. /2 2.3 Next, integrate (cos(t))2n+1dt and show that it can be written as 2n I(n) = I(n - 1) 2n + 1 Hint: Integrate by parts. Split the exponent so that dv = cos(t)dt 3. Now that we have derived I(n) = 22(n-1), and since I(1) = 4/3 3.1 Confirm I(2)= 15 by using the new formula I(n) 16 2n 2n+1 (n - 1) 4. If we calculate I(3) using I(n) = 211(n - 1), we observe that 1(3) = -1 (2) I 3.3.2, and since I(2) = 16 or 2, then 642 - I(3) = 7 5 3 2 (This is known as the Wallis Product) Using I(n) =22I(n - 1) formula, calculate: 4.1 I(4) 4.2 I(5) (For 4.1 and 4.2, leave your answers in fractional format) 5. You should have observed that the general pattern using the fractional format is: I(n) = 2 2n 2n+1 8642 9 7 53 5.1 Use this fractional pattern to show that: I(n) = 22n+1 (n!)² (2n+1)! Recall that n! = n (n-1) (n-2)-4-3.2.1 Hint: Multiply the fractional pattern by 8 20100 8 2n and modify 2n the newly formed expression to its final form.
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Transcribed Image Text:Given the following integral I(n) = (1-2)"dx. We want to ultimately
show that
22n+1 (n!)2
I(n) =
(2n+1)!
I
1. Use I(n) = f₁(1 − x²)" dr to evaluate:
1.1 I(1)
1.2 I(2)
2. Recall that I(n) = f₁(1 - x²)" dx.
2.1 Substitute x = sin(t) in I(n) and show that I(n) can be written as
π/2
L
-*/2
(cos(t))2n+1dt
I(n) =
2.2 Now that we have I(n) =
1/2
(cos(t))2n+1dt. What is I(n - 1)?
Leave your answer in integral form.
Transcribed Image Text:/2
2.3 Next, integrate (cos(t))2n+1dt and show that it can be written
as
2n
I(n) =
I(n - 1)
2n + 1
Hint: Integrate by parts. Split the exponent so that dv = cos(t)dt
3. Now that we have derived I(n) = 22(n-1), and since I(1) = 4/3
3.1 Confirm I(2)= 15 by using the new formula I(n)
16
2n
2n+1 (n - 1)
4. If we calculate I(3) using I(n) = 211(n - 1), we observe that
1(3) = -1 (2)
I
3.3.2,
and since I(2) = 16 or 2, then
642
-
I(3) = 7 5 3
2 (This is known as the Wallis Product)
Using I(n) =22I(n - 1) formula, calculate:
4.1 I(4)
4.2 I(5)
(For 4.1 and 4.2, leave your answers in fractional format)
5. You should have observed that the general pattern using the fractional
format is:
I(n) = 2
2n
2n+1
8642
9 7 53
5.1 Use this fractional pattern to show that:
I(n) =
22n+1 (n!)²
(2n+1)!
Recall that n! = n (n-1) (n-2)-4-3.2.1
Hint: Multiply the fractional pattern by
8
20100
8 2n and modify
2n
the newly formed expression to its final form.
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