Complex! Yet Lollipop!

Calculus Level 3

Evaluate

⌊ ∫ 3 99 ( 1 + x ) [ ( 1 − x + x 2 ) ( 1 + x + x 2 ) + x 2 ] 1 + 2 x + 3 x 2 + 4 x 3 + 3 x 4 + 2 x 5 + x 6 d x ⌋ . \lfloor \int_3^{99} { \frac { \left( 1 + x \right) \left[ \left( 1 - x + { x }^{ 2 } \right) \left( 1 + x + { x }^{ 2 } \right) + { x }^{ 2 } \right] }{ 1 + 2x + 3{ x }^{ 2 } + 4{ x }^{ 3 } + 3{ x }^{ 4 } + 2{ x }^{ 5 } + { x }^{ 6 } } dx } \rfloor.

Notation : ⌊ ⋅ ⌋ \lfloor \cdot \rfloor denotes the floor function .


The answer is 3.

Priyanshu Mishra by Priyanshu Mishra , 20, India

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1 solution

Chew-Seong Cheong
Jun 26, 2017

F ( 99 ) − F ( 3 ) = ∫ 3 99 ( 1 + x ) [ ( 1 − x + x 2 ) ( 1 + x + x 2 ) + x 2 ] 1 + 2 x + 3 x 2 + 4 x 3 + 3 x 4 + 2 x 5 + x 6 d x = ∫ 3 99 ( 1 + x ) [ ( 1 + x 2 − x ) ( 1 + x 2 + x ) + x 2 ] 1 + 2 x + x 2 + 2 x 2 + 4 x 3 + 2 x 4 + x 4 + 2 x 5 + x 6 d x = ∫ 3 99 ( 1 + x ) [ ( 1 + x 2 ) 2 − x 2 + x 2 ] ( 1 + x ) 2 + 2 x 2 ( 1 + x ) 2 + x 4 ( 1 + x ) 2 d x = ∫ 3 99 ( 1 + x ) ( 1 + x 2 ) 2 ( 1 + x ) 2 ( 1 + 2 x 2 + x 4 ) d x = ∫ 3 99 ( 1 + x ) ( 1 + x 2 ) 2 ( 1 + x ) 2 ( 1 + x 2 ) 2 d x = ∫ 3 99 1 1 + x d x = ln ⁡ ( 1 + x ) ∣ 3 99 = ln ⁡ 100 − ln ⁡ 4 = ln ⁡ 25 ≈ 3.2189 \begin{aligned} F(99) - F(3) & = \int_3^{99} \frac {(1+x)\left[\left(1-x+x^2\right) \left(1+x+x^2\right)+ x^2 \right]}{1+2x+{\color{#3D99F6}3x^2} +4x^3+{\color{#3D99F6}3x^4}+2x^5+x^6} dx \\ & = \int_3^{99} \frac {(1+x)\left[\left({\color{#3D99F6}1+x^2}{\color{#D61F06}-x}\right) \left({\color{#3D99F6}1+x^2}{\color{#D61F06}+x}\right)+ x^2 \right]}{1+2x+{\color{#3D99F6}x^2 +2x^2}+4x^3+{\color{#3D99F6}2x^4+x^4}+2x^5+x^6} dx \\ & = \int_3^{99} \frac {(1+x)\left[{\color{#3D99F6}(1+x^2)^2}{\color{#D61F06}-x^2}+ x^2 \right]}{(1+x)^2+2x^2(1+x)^2+x^4(1+x)^2} dx \\ & = \int_3^{99} \frac {(1+x)(1+x^2)^2}{(1+x)^2(1+2x^2+x^4)} dx \\ & = \int_3^{99} \frac {(1+x)(1+x^2)^2}{(1+x)^2(1+x^2)^2} dx \\ & = \int_3^{99} \frac 1{1+x} dx \\ & = \ln (1+x) \bigg|_3^{99} = \ln 100 - \ln 4 = \ln 25 \approx 3.2189 \end{aligned}

⟹ ⌊ F ( 99 ) − F ( 3 ) ⌋ = 3 \implies \lfloor F(99) - F(3) \rfloor = \boxed{3}

2 Helpful 0 Interesting 1 Brilliant 0 Confused

@Chew-Seong Cheong , Really class solution sir. Mine was very lengthy.

Priyanshu Mishra - 3 years, 11 months ago

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