Limit, sum and integration!

Calculus Level 3

lim ⁡ n → ∞ 1 n 2 ∑ k = 1 n − 1 ( k ∫ k k + 1 ( x − k ) ( k + 1 − x ) d x ) \lim_{n\to \infty} \dfrac{1}{n^2} \sum_{k=1}^{n-1}\left(k \int_{k}^{k+1} \sqrt{\,(x-k)\,(k+1-x)} \,dx\right) If the value of above expression can be written as π b \dfrac{\pi}{b} , where b b is positive integer. Find the value of 2 b 2b .

Source: Sagar Kumar

4 16 32 8

Naren Bhandari by Naren Bhandari , 21, India

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1 solution

Naren Bhandari
Aug 10, 2018

My Approach ζ = lim ⁡ n → ∞ 1 n 2 ∑ k = 0 n − 1 ( k ∫ k k + 1 ( x − k ) ( k + 1 − x ) d x ) \zeta = \lim_{n\to\infty} \dfrac{1}{n^2} \sum_{k=0}^{n-1}\left(k\int_{k}^{k+1} \sqrt{(x-k)(k+1-x)} \,dx\right) Let us determine the integral of ∫ k k + 1 ( x − k ) ( k + 1 − x ) d x \int_{k}^{k+1} \sqrt{(x-k)(k+1-x)}\,dx Making u u substitution of k + 1 − x = u ⇒ \sqrt{k+1-x } = u \Rightarrow u 2 = k + 1 − x ⟹ x = − u 2 + k + 1 u^2 = k+1-x \implies x = -u^2+k+1 .Thus, d x = − 2 u d u \,dx = -2u\,du . Now ∫ 1 0 − 2 u 2 1 − u 2 d u = − 2 ∫ 1 0 ( u 2 − 1 + 1 ) 1 − u 2 d u = ∫ 1 0 ( 1 − u 2 − ( 1 − u 2 ) 3 2 ) d u \begin{aligned} \int_{1}^{0} - 2u^2\sqrt{1-u^2}\,du & = - 2\int_{1}^{0} \,(u^2-1+1)\sqrt{1-u^2}\,du \\ &= \int_{1}^{0} \left(\sqrt{1-u^2} -{\,(1-u^2)}^{\frac{3}{2}}\right) \,du \end{aligned} Here again we make substitution u = sin ⁡ ϕ u=\sin \phi ⟹ d u = cos ⁡ ϕ \implies \,du = \cos \phi .Pluging back to integral we have ∫ 1 0 ( 1 − u 2 − ( 1 − u 2 ) 3 2 ) d u = ∫ 0 π 2 ( cos ⁡ 2 ϕ − cos ⁡ 4 ϕ ) d ϕ = ∫ 0 π 2 { ( e i ϕ + e − i ϕ 2 ) 2 − ( e i ϕ + e − i ϕ 2 ) 4 } d ϕ = ∫ 0 π 2 { 1 + cos ⁡ ϕ 2 − cos ⁡ 4 ϕ + 4 cos ⁡ 2 ϕ + 3 8 } d ϕ = 2 ϕ + sin ⁡ 2 ϕ 4 − sin ⁡ 4 ϕ + 8 sin ⁡ sin ⁡ 2 ϕ + 12 ϕ 32 ∣ 0 π 2 \begin{aligned} \int_{1}^{0} \left(\sqrt{1-u^2} -{\,(1-u^2)}^{\frac{3}{2}}\right) \,du & = \int_{0}^{\frac{\pi}{2}} \left(\cos^2 \phi - \cos ^4 \phi \right) \,d\phi \\ & = \int_{0}^{\frac{\pi}{2}} \left\{\left(\dfrac{e^{i\phi} + e^{-i\phi}}{2}\right)^2-\left(\dfrac{e^{i\phi} + e^{-i\phi}}{2}\right)^4\right\}\,d\phi \\& = \int_{0}^{\frac{\pi}{2}}\left\{\dfrac{1+\cos \phi}{2} - \dfrac{\cos 4\phi +4\cos 2\phi +3}{8} \right\}\,d\phi \\& = \dfrac{2\phi + \sin 2\phi }{4}- \dfrac{\sin 4 \phi + 8 \sin \sin 2\phi +12 \phi }{32}\large {|}_{0}^{\frac{\pi}{2}}\end{aligned} Now setting the upper and lower limit we find ∫ 1 0 ( 1 − u 2 − ( 1 − u 2 ) 3 2 ) d u = − π 16 \begin{aligned} \int_{1}^{0} \left(\sqrt{1-u^2} -{\,(1-u^2)}^{\frac{3}{2}}\right) \,du =- \dfrac{\pi}{16} \end{aligned} Therefore, − 2 ∫ 1 0 ( 1 − u 2 − ( 1 − u 2 ) 3 2 ) d u = π 8 \begin{aligned}-2 \int_{1}^{0} \left(\sqrt{1-u^2} -{\,(1-u^2)}^{\frac{3}{2}}\right) \,du= \dfrac{\pi}{8}\end{aligned} Heading back to ζ = lim ⁡ n → ∞ 1 n 2 ∑ k = 1 n − 1 π k 8 = π 8 ( lim ⁡ n → ∞ 1 n ∑ k = 1 n k n ) c c c c Riemann sum = π 8 ( ∫ 0 1 x d x ) = π 8 ( x 2 2 ∣ 0 1 ) = π 16 \begin{aligned} \zeta & = \lim_{n\to \infty} \dfrac{1}{n^2} \sum_{k=1}^{n-1} \dfrac{\pi k}{8} = \dfrac{\pi}{8}\left(\lim_{n\to \infty} \dfrac{1}{n}\sum_{k=1}^{n}\dfrac{k}{n}\right)\phantom{cccc}{\color{#3D99F6}\text{Riemann sum}} \\& = \dfrac{\pi}{8} \left(\int_{0}^{1} x\,dx\right) = \dfrac{\pi}{8} \left(\dfrac{x^2}{2}\large{|}_{0}^{1}\right) =\boxed{ \dfrac{\pi}{16}}\end{aligned} Therefore, 2 b = 32 2b=32 .

4 Helpful 0 Interesting 1 Brilliant 0 Confused

I missed the " 2 b 2b " part, so can't post my solution, but I wanted to point out that if you start with ∫ k k + 1 ( x − k ) ( k + 1 − x ) d x \int_{k}^{k+1} \sqrt{(x-k)(k+1-x)}\,dx and just apply the substitution u = x − k − 1 2 , u = x-k-\frac{1}{2}, then it becomes ∫ − 1 / 2 1 / 2 ( 1 2 ) 2 − u 2 d u \int_{-1/2}^{1/2} \sqrt{\left(\frac{1}{2}\right)^2 - u^2}\,du which you can now recognize as half the area of a circle with radius 1 2 , \frac{1}{2}, or 1 2 π ( 1 2 ) 2 = π 8 \frac{1}{2} \pi \left(\frac{1}{2}\right)^2 = \frac{\pi}{8}

Brian Moehring - 2 years, 10 months ago

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Oh!! It's sad to learn that. I'm glad to go through your solution which is pretty short and nice. Thank you for the solution .

Naren Bhandari - 2 years, 10 months ago

I would like to know, "What things we must care while solving integrations problems?" Like how we used observe the problem and how we should analysis it and take wise steps to begin the problem. I would be grateful if you could provide me any such sorts of tricks or ways to deal with problems. :)

Naren Bhandari - 2 years, 10 months ago

I was tricked by the " 2 b 2b " part. Don't set a question like that especially a multiple choice one. You are checking the avoiding trick skill other than calculus. Answer option is to avoid the use of calculator to answer the problem. The d x dx should be within the bracket not outside.

Chew-Seong Cheong - 2 years, 10 months ago

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I have a simpler solution than yours. Now I cannot show it.

Chew-Seong Cheong - 2 years, 10 months ago

L = lim ⁡ n → ∞ 1 n 2 ∑ k = 1 n − 1 k ∫ k k + 1 ( x − k ) ( k + 1 − x ) d x Let u = x − k ⟹ d u = d x = lim ⁡ n → ∞ 1 n 2 ∑ k = 1 n − 1 k ∫ 0 1 u ( 1 − u ) d u Let u = sin ⁡ 2 θ ⟹ d u = 2 sin ⁡ θ cos ⁡ θ d θ = lim ⁡ n → ∞ 1 n 2 ∑ k = 1 n − 1 k ∫ 0 π 2 2 sin ⁡ 2 θ cos ⁡ 2 θ d θ = lim ⁡ n → ∞ 1 n 2 ∑ k = 1 n − 1 k ∫ 0 π 2 sin ⁡ 2 2 θ 2 d θ = lim ⁡ n → ∞ 1 n 2 ∑ k = 1 n − 1 k 4 ∫ 0 π 2 ( 1 − cos ⁡ 4 θ ) d θ = lim ⁡ n → ∞ 1 n 2 ∑ k = 1 n − 1 k 4 [ θ − sin ⁡ 4 θ 4 ] 0 π 2 = lim ⁡ n → ∞ 1 n 2 ∑ k = 1 n − 1 k π 8 = lim ⁡ n → ∞ n ( n − 1 ) π 16 n 2 Divide up and down with n 2 = lim ⁡ n → ∞ ( 1 − 1 n ) π 16 = π 16 \begin{aligned} L & = \lim_{n \to \infty} \frac 1{n^2} \sum_{k=1}^{n-1} k \int_k^{k+1} \sqrt{(x-k)(k+1-x)} dx & \small \color{#3D99F6} \text{Let }u = x-k \implies du = dx \\ & = \lim_{n \to \infty} \frac 1{n^2} \sum_{k=1}^{n-1} k \int_0^1 \sqrt{u(1-u)} du & \small \color{#3D99F6} \text{Let }u = \sin^2 \theta \implies du = 2\sin \theta \cos \theta \ d\theta \\ & = \lim_{n \to \infty} \frac 1{n^2} \sum_{k=1}^{n-1} k \int_0^\frac \pi 2 2\sin^2 \theta \cos^2 \theta \ d\theta \\ & = \lim_{n \to \infty} \frac 1{n^2} \sum_{k=1}^{n-1} k \int_0^\frac \pi 2 \frac {\sin^2 2\theta}2 \ d\theta \\ & = \lim_{n \to \infty} \frac 1{n^2} \sum_{k=1}^{n-1} \frac k4 \int_0^\frac \pi 2 (1-\cos 4\theta)\ d\theta \\ & = \lim_{n \to \infty} \frac 1{n^2} \sum_{k=1}^{n-1} \frac k4 \left[\theta - \frac {\sin 4\theta}4\right]_0^\frac \pi 2 \\ & = \lim_{n \to \infty} \frac 1{n^2} \sum_{k=1}^{n-1} \frac {k\pi}8 \\ & = \lim_{n \to \infty} \frac {n(n-1)\pi}{16n^2} & \small \color{#3D99F6} \text{Divide up and down with }n^2 \\ & = \lim_{n \to \infty} \frac {\left(1-\frac 1n\right)\pi}{16} \\ & = \frac \pi{16} \end{aligned}

Therefore, 2 b = 2 ( 16 ) = 32 2b = 2(16) = \boxed{32} .

Chew-Seong Cheong - 2 years, 10 months ago

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