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Problem 74 Hard Difficulty

Let $ I_n = \displaystyle \int_0^{\frac{\pi}{2}} \sin^n x dx $.
(a) Show that $ I_{2n + 2} \le I_{2n + 1} \le I_{2n} $.
(b) Use Exercise 50 to show that
$$ \frac{I_{2n + 2}}{I_{2n}} = \frac{2n + 1}{2n + 2} $$
(c) Use parts (a) and (b) to show that
$$ \frac{2n + 1}{2n + 2} \le \frac{I_{2n + 1}}{I_{2n}} \le 1 $$
and deduce that $ \lim_{n\to\infty}\frac{I_{2n + 1}}{I_{2n}} = 1 $.
(d) Use part (c) and Exercises 49 and 50 to show that
$$ \lim_{n\to\infty} \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdots \cdots \frac{2n}{2n - 1} \cdot \frac{2n}{2n + 1} = \frac{\pi}{2} $$
This formula is usually written as an infinite product:
$$ \frac{\pi}{2} = \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdots $$
and is called the Wallis product.
(e) We construct rectangles as follows. Start with a square of area 1 and attach rectangles of area 1 alternately beside or on top of the previous rectangle (see the figure). Find the limit of the ratios of width to height of these rectangles.

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Video Transcript

The problem is that a m is equal to integral from 0 to pi over 2 sine x to n power. Dx is shown that i 2 n, plus 2 is smaller than i 2 n plus 1 is smaller than i to the reason is for a the reason is sine x is smaller than land greater than 0 point, so they have a sine 2 n plus 2 X is smaller than sine 2 n. Plus 1 x is smaller than sine 2 n x, so they have. I 2 plus 2, is smaller than i to plus 1 is smaller than i 2 in b is use exercise 50 to show that i 2 m plus 2 over i 2 n is equal to 2 n plus 1 over 2 plus 2 t in exercise 50. We half integral from 0 to pi over 2 sine x to x power dis equal to 1 over n integral from 0 to pi over 2 sine x to minus las power, dx so replace in by 2, n plus 2, and they have this result. For part c, ives part a part b to show that i 2 n plus 1 over 2, is greater than 2 n plus 1 over 2 plus 2 n smaller than 1 point. For this part, this is a director result from a, and for this part we can see, i 2 n plus 1. Over i, 2 n is greater than i 2 n plus 2 over i 2, by a and then by. This is just 2 n plus 1. Over 2 n plus 2 point and by square serum is limit of this. 1 is equal to 1 goes to infinity, and this is also 1 to the limit of this. 1 is equal to 1, but do need to prove the limit of this number is equal to pi over 2 point here for simplicity. We write this number as a a n witses an or i can write. This is a 2 n. So then, what angles infinity we need to prove. The limit of a 2 n is equal to pi over 2. By the result of an exercise 49 point we have, i 2 m, plus 1 is equal to 2 times 4 times 6 times times: 2 n over 3 times 5 times 7 times, 2 n plus 1, and i by exercise 50 and 49. We have a 2 n is equal to 1 times 3 times 5 times, dot, dot 2 m minus 1 over 2 times 4 times to the dod times 2 n times pi over 2. So we can find i 2 n plus 1 over in is just a 2 n times 2 over pi. Then this as a result of a limit and goes to infinity, i 2 n plus 1 over i i 2 n 2 n. This is equal to 1. So we have the limit of a 2 n is equal to p o 2 for the last part for e. Let'S look at some rectangles. We construct rectangles as follows: start with a square of ire 1 and attach rectangles of area 1 alternately beside on top of the previous rectangles, for example. This is 111 and third graph. Is this 1 just 1 kind of top things this 1 on 1? So this is 1 half and is the force along this from here we at rectangles here to this 1111 half- and here this is 2 over 3 and fifth 1 is like- and this is 1 half 1112 over 3. This is 8 over 3. So this this side, so this is 3 over 8. So what can the question is find the limit of the ratio of is to a height of these rectangles so for 1. The first 1 is the first 1 that reaches is over height is equal to 1. Second, 1, is over height is equal to 2. Over 1 point. Third, 1 is wise: over height is equal to 2 over 1 plus 1 half pint. This is equal to 2 over 3 over 2. So this 4 over 3 point. So we can write that this is 2 over 1 times 2 over 3 and the fourth 1 was over. Heit is equal to 8 over 3 over 3. Over 2 point. This is 16 o 9. So we can write that this is 2 over 1 times 2. Over 3 times 4 over 3 and the fifth is lost over height. This is 8 over 3 over 1 plus ne half plus 3 over 8 point, and you can check this is equal to 2 over 1 times 2 over 3 times 4 over 3 times 4. Over 5 point so just to compare this result with part d you can see. This is exactly the number of a 2 notely know that limit is equal to pi over 2 by the result of part b.

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James Stewart

Calculus: Early Transcendentals

Grace He

Samuel Hannah

Michael Jacobsen

Joseph Lentino

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Grace He

Samuel Hannah

Michael Jacobsen

Joseph Lentino

James StewartCalculus: Early Transcendentals

Grace He

Samuel Hannah

Michael Jacobsen

Joseph Lentino

James Stewart

Calculus: Early Transcendentals