"Useful" Integration

Calculus Level 3

A = ∫ 0 1 x 4 ( 1 − x ) 4 1 + x 2 d x \large A = \int_0^1 \frac{x^4(1-x)^4}{1+x^2} dx

For A A is defined above, find ⌊ 1 0 6 A ⌋ \left \lfloor 10^6A\right \rfloor ?

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

Bonus: Which well-known assumption does this problem disprove?


The answer is 1264.

Krishnaraj Sambath by Krishnaraj Sambath , 35, USA

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1 solution

Chew-Seong Cheong
Jan 18, 2019

A = ∫ 0 1 x 4 ( 1 − x ) 4 1 + x 2 d x = ∫ 0 1 x 8 − 4 x 7 + 6 x 6 − 4 x 5 + x 4 x 2 + 1 d x = ∫ 0 1 ( x 6 − 4 x 5 + 5 x 4 − 4 x 2 + 4 − 4 x 2 + 1 ) d x = x 7 7 − 4 6 x 6 + x 5 − 4 3 x 3 + 4 x − 4 tan ⁡ − 1 x ∣ 0 1 = 22 7 − π ≈ 0.001264489267349614 \begin{aligned} A & = \int_0^1 \frac {x^4(1-x)^4}{1+x^2} dx \\ & = \int_0^1 \frac {x^8-4x^7+6x^6-4x^5+x^4}{x^2+1} dx \\ & = \int_0^1 \left(x^6-4x^5+5x^4-4x^2+4-\frac 4{x^2+1}\right) dx \\ & = \frac {x^7}7 - \frac 46x^6 + x^5 - \frac 43x^3 + 4x - 4\tan^{-1}x \ \bigg|_0^1 \\ & = \frac {22}7 - \pi \\ & \approx 0.001264489267349614 \end{aligned}

Therefore ⌊ 1 0 6 A ⌋ = 1264 \left \lfloor 10^6 A \right \rfloor = \boxed{1264} .

Bonus: This proves that π ≠ 22 7 \pi \ne \frac {22}7 .

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