The integral that was famous

Calculus Level 3

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

The value of the integral I I is

A. 2 105 \quad \text{A.} \quad \dfrac{2}{105}

B. 0 \quad \text{B.} \quad 0

C. 22 7 − π \quad \text{C.} \quad \dfrac{22}{7} - \pi

D. π − 22 7 \quad \text{D.} \quad \pi - \dfrac{22}{7}

E. 71 115 + 3 π 2 \quad \text{E.} \quad \dfrac{71}{115} +\dfrac{3\pi}{2}

D A B C E

Ayon Ghosh by Ayon Ghosh , 17, India

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1 solution

Chew-Seong Cheong
Jul 23, 2017

I = ∫ 0 1 x 4 ( 1 − x ) 4 1 + x 2 d x = ∫ 0 1 x 4 ( x − 1 ) 4 x 2 + 1 d x = ∫ 0 1 x 4 ( x 4 − 4 x 3 + 6 x 2 − 4 x + 1 ) x 2 + 1 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 8 + x 6 − 4 x 7 − 4 x 5 + 5 x 6 + 5 x 4 − 4 x 4 − 4 x 2 + 4 x 2 + 4 − 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 x 6 6 + 5 x 5 5 − 4 x 3 3 + 4 x − 4 tan ⁡ − 1 x ] 0 1 = 1 7 − 2 3 + 1 − 4 3 + 4 − 4 × π 4 = 22 7 − π \begin{aligned} I & = \int_0^1 \frac {x^4(1-x)^4}{1+x^2} \ dx \\ & = \int_0^1 \frac {x^4(x-1)^4}{x^2+1} \ dx \\ & = \int_0^1 \frac {x^4(x^4-4x^3 +6 x^2-4x+1)}{x^2+1} \ dx \\ & = \int_0^1 \frac {x^8-4x^7 +6x^6-4x^5+x^4}{x^2+1} \ dx \\ & = \int_0^1 \frac {x^8+x^6-4x^7-4x^5 +5x^6+5x^4-4x^4-4x^2+4x^2+4-4}{x^2+1} \ dx \\ & = \int_0^1 \left(x^6-4x^5+5x^4-4x^2+4- \frac 4{x^2+1} \right) dx \\ & = \left[ \frac {x^7}7 - \frac {4x^6}6 + \frac {5x^5}5 - \frac {4x^3}3 + 4x - 4\tan^{-1} x \right]_0^1 \\ & = \frac 17 - \frac 23 + 1 - \frac 43 + 4 - 4\times \frac \pi 4 \\ & = \boxed{\dfrac {22}7-\pi} \end{aligned}

Therefore, the answer is C .

4 Helpful 0 Interesting 0 Brilliant 0 Confused

This is famously the Putnam competition proof that 22 7 > π \dfrac{22}{7} > \pi , since we know that the integrand and therefore the integral is always positive.

Zach Abueg - 3 years, 10 months ago

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Yes, seen this before. But not very sure about it. Thanks.

Chew-Seong Cheong - 3 years, 10 months ago

yes that's what the integral is famous for.

Ayon Ghosh - 3 years, 10 months ago

it had also appeared in the IIT-JEE exam of 2010.

Ayon Ghosh - 3 years, 10 months ago

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