Let's insert 1729 again and again

Calculus Level 5

Given : ∫ 0 2 × 1729 π max ⁡ { sin ⁡ x , sin ⁡ − 1 ( sin ⁡ x ) } d x = a \int_{0}^{2 \times 1729\pi} \max\{\sin x, \sin^{-1}(\sin x)\} \ dx=a

Find the value of ⌈ a ⌉ \lceil a \rceil

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Note : ⌈ . . ⌉ \lceil .. \rceil denotes the Ceiling Function( lowest integer greater than equal to).


The answer is 809.

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Sandeep Bhardwaj by Sandeep Bhardwaj , 28, India

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1 solution

Ayush Verma
Nov 20, 2014

B o t h f u n c t i o n h a v e p e r i o d 2 π , s o a = 1729 ∫ 0 2 π m a x ( { sin ⁡ x , sin ⁡ − 1 ( sin ⁡ x ) } ) d x = 1729 × I m a x ( { sin ⁡ x , sin ⁡ − 1 ( sin ⁡ x ) } ) = sin ⁡ − 1 ( sin ⁡ x ) , x ∈ ( 0 , π ) m a x ( { sin ⁡ x , sin ⁡ − 1 ( sin ⁡ x ) } ) = sin ⁡ x , x ∈ ( π , 2 π ) I = ∫ 0 π sin ⁡ − 1 ( sin ⁡ x ) d x + ∫ π 2 π s i n x d x = I 1 + I 2 I 2 = − 2 I 1 = A r e a o f △ f o r m e d b y y = 0 , y = x & y = π − x = 1 2 × π × π 2 = π 2 4 ∴ a = 1729 × I = 1729 ( π 2 4 − 2 ) = 808.13 ⌈ a ⌉ = 809 Both\quad function\quad have\quad period\quad 2\pi ,so\\ \\ a=1729\int _{ 0 }^{ 2\pi }{ max\left( \left\{ \sin { x,\sin ^{ -1 }{ \left( \sin { x } \right) } } \right\} \right) dx } =1729\times I\\ \\ \\ max\left( \left\{ \sin { x,\sin ^{ -1 }{ \left( \sin { x } \right) } } \right\} \right) =\sin ^{ -1 }{ \left( \sin { x } \right) } ,\quad x\in \left( 0,\pi \right) \\ \\ max\left( \left\{ \sin { x,\sin ^{ -1 }{ \left( \sin { x } \right) } } \right\} \right) =\sin { x } ,\quad x\in \left( \pi ,2\pi \right) \\ \\ I=\int _{ 0 }^{ \pi }{ \sin ^{ -1 }{ \left( \sin { x } \right) } dx } +\int _{ \pi }^{ 2\pi }{ sinxdx } ={ I }_{ 1 }+{ I }_{ 2 }\\ \\ { I }_{ 2 }=-2\\ \\ { I }_{ 1 }=Area\quad of\quad \triangle \quad formed\quad by\quad y=0,y=x\quad \& \quad y=\pi -x\\ \\ \quad =\cfrac { 1 }{ 2 } \times \pi \times \cfrac { \pi }{ 2 } =\cfrac { { \pi }^{ 2 } }{ 4 } \\ \\ \therefore a=1729\times I=1729\left( \cfrac { { \pi }^{ 2 } }{ 4 } -2 \right) =808.13\\ \\ \left\lceil a \right\rceil =809\\

15 Helpful 0 Interesting 0 Brilliant 0 Confused

Easy but really interesting question.

Jake Lai - 6 years, 6 months ago

I am getting 803.4 by same method

aryaman jha - 5 years, 11 months ago

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may be calculation mistake.

use arcsin(sinx)=x for 2x=[0 , pi}

& arcsin(sinx)=pi-x for 2x=[pi , 2pi}

Ayush Verma - 5 years, 11 months ago

I mean, we used the exact same method. Even the steps! High Five!

Kishore S. Shenoy - 4 years, 8 months ago

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