Stuffed with everything I could think of

Calculus Level 2

f ( x ) = ∣ s i n 5 x l n ( s i n ( x ) ) s i n ( x ) s i n ( x ) + c o s ( x ) ⌈ n ⌉ ⌊ ∑ k = 1 10000 k ⌋ { ∏ k = 1 10000 k } 8 15 π 2 l n ( 1 2 ) π 4 ∣ \large{f\left( x \right) =\left| \begin{matrix} { sin }^{ 5 }x\quad & ln\left( sin(x) \right) \quad & \frac { \sqrt { sin(x) } }{ \sqrt { sin(x) } +\sqrt { cos(x) } } \\ \left\lceil n \right\rceil & \left\lfloor \sum _{ k=1 }^{ 10000 }{ k } \right\rfloor & \left\{ \prod _{ k=1 }^{ 10000 }{ k } \right\} \\ \frac { 8 }{ 15 } & \frac { \pi }{ 2 } ln\left( \frac { 1 }{ 2 } \right) & \frac { \pi }{ 4 } \end{matrix} \right| }

∴ { ∫ 0 π 2 f ( x ) d x } = ? \large{\therefore \quad \quad \quad \left\{ \int _{ 0 }^{ \frac { \pi }{ 2 } }{ f\left( x \right) } dx \right\} =?}

⌈ . ⌉ a n d ⌊ . ⌋ \left\lceil . \right\rceil \quad and\quad \left\lfloor . \right\rfloor represents the ceiling and floor function respectively.

{ . } \left\{ . \right\} is the fractional part function.

∑ a n d ∏ \sum { } \quad and\quad \prod { } are sum and product functions respectively.

This is a problem of my set JEE Calculus .


The answer is 0.

Harshvardhan Mehta by Harshvardhan Mehta , 23, India

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1 solution

∫ 0 π 2 f ( x ) d x = ∣ ∫ 0 π 2 s i n 5 x d x − ∫ 0 π 2 l n ( s i n ( x ) ) d x ∫ 0 π 2 s i n ( x ) s i n ( x ) + c o s ( x ) d x ⌈ n ⌉ ⌊ ∑ k = 1 10000 k ⌋ { ∏ k = 1 10000 k } 8 15 π 2 l n ( 1 2 ) π 4 ∣ \int _{ 0 }^{ \frac { \pi }{ 2 } }{ f\left( x \right) } dx=\left| \begin{matrix} \int _{ 0 }^{ \frac { \pi }{ 2 } }{ { sin }^{ 5 }x } dx\quad & -\int _{ 0 }^{ \frac { \pi }{ 2 } }{ ln\left( sin(x) \right) } dx\quad & \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \sqrt { sin(x) } }{ \sqrt { sin(x) } +\sqrt { cos(x) } } } dx \\ \left\lceil n \right\rceil & \left\lfloor \sum _{ k=1 }^{ 10000 }{ k } \right\rfloor & \left\{ \prod _{ k=1 }^{ 10000 }{ k } \right\} \\ \frac { 8 }{ 15 } & \frac { \pi }{ 2 } ln\left( \frac { 1 }{ 2 } \right) & \frac { \pi }{ 4 } \end{matrix} \right|

= ∣ 8 15 − π 2 l n ( 2 ) π 4 ⌈ n ⌉ ⌊ ∑ k = 1 10000 k ⌋ { ∏ k = 1 10000 k } 8 15 π 2 l n ( 1 2 ) π 4 ∣ =\left| \begin{matrix} \frac { 8 }{ 15 } \quad & -\frac { \pi }{ 2 } ln\left( 2 \right) \quad & \frac { \pi }{ 4 } \\ \left\lceil n \right\rceil & \left\lfloor \sum _{ k=1 }^{ 10000 }{ k } \right\rfloor & \left\{ \prod _{ k=1 }^{ 10000 }{ k } \right\} \\ \frac { 8 }{ 15 } & \frac { \pi }{ 2 } ln\left( \frac { 1 }{ 2 } \right) & \frac { \pi }{ 4 } \end{matrix} \right|

= ∣ 8 15 π 2 l n ( 1 2 ) π 4 ⌈ n ⌉ ⌊ ∑ k = 1 10000 k ⌋ { ∏ k = 1 10000 k } 8 15 π 2 l n ( 1 2 ) π 4 ∣ =\left| \begin{matrix} \frac { 8 }{ 15 } \quad & \frac { \pi }{ 2 } ln\left( \frac { 1 }{ 2 } \right) \quad & \frac { \pi }{ 4 } \\ \left\lceil n \right\rceil & \left\lfloor \sum _{ k=1 }^{ 10000 }{ k } \right\rfloor & \left\{ \prod _{ k=1 }^{ 10000 }{ k } \right\} \\ \frac { 8 }{ 15 } & \frac { \pi }{ 2 } ln\left( \frac { 1 }{ 2 } \right) & \frac { \pi }{ 4 } \end{matrix} \right|

S i n c e R 1 a n d R 3 a r e i d e n t i c a l . Since\quad R_{ 1 }\quad and\quad R_{ 3 }\quad are\quad identical.

∴ ∫ 0 π 2 f ( x ) d x = 0 \therefore \quad \quad \quad \quad \quad \int _{ 0 }^{ \frac { \pi }{ 2 } }{ f\left( x \right) } dx\quad =\quad 0

⇒ { ∫ 0 π 2 f ( x ) d x } = 0 \Rightarrow \quad \quad \quad \quad \quad \left\{ \int _{ 0 }^{ \frac { \pi }{ 2 } }{ f\left( x \right) } dx \right\} \quad =\quad \boxed{0}

8 Helpful 1 Interesting 1 Brilliant 0 Confused

Haha ! Looked pretty scary at the first glance. ! : ) :)

Keshav Tiwari - 6 years, 1 month ago

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Haha.. That was the main objective.. :P ;)

Harshvardhan Mehta - 6 years, 1 month ago

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