Fun with Integration

Calculus Level 2

∫ 0 1 ( 1 − x 2 ) 9 x 9 d x \int_0^1 \left(1-x^2\right)^9 x^9 \, dx

Let I I denote the value of the integral above. What is the sum of digits of I − 1 ? I^{-1}?


The answer is 4.

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Aritra Kundu by Aritra Kundu , 23, India

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5 solutions

Tanishq Varshney
Jul 25, 2015

let t = x 2 t=x^2

d t = 2 x d x dt=2x dx

∫ 0 1 1 2 ( 1 − t ) 9 t 4 d t \large{\displaystyle \int^{1}_{0} \frac{1}{2}(1-t)^{9}t^4dt}

using beta function

∫ 0 1 ( 1 − t ) x − 1 t y − 1 = ( x − 1 ) ! ( y − 1 ) ! ( x + y − 1 ) ! \large{\displaystyle \int^{1}_{0} (1-t)^{x-1}t^{y-1}=\frac{(x-1)! (y-1)!}{(x+y-1)!}}

I = 1 2 × 9 ! × 4 ! 14 ! I=\frac{1}{2} \times \frac{9! \times 4!}{14! }

I − 1 = 20020 I^{-1}=20020

46 Helpful 2 Interesting 10 Brilliant 1 Confused

i miscalculated twice and the third time when i got the right answer i missed that the answer entered should be sum of digits of the inverse of I . silly me.

Anurag Pandey - 3 years ago
Aritra Kundu
Jul 25, 2015

21 Helpful 3 Interesting 2 Brilliant 0 Confused

Moderator note:

You don't need to use trigonometric substitution. A better substitution to use is y = 1 − x 2 y = 1-x^2 or y = x 2 y=x^2 . Do you see why?

And there's a simpler approach afterwards too. Are you familiar with Beta functions?

∫ 0 1 x m ( 1 − x ) n d x = m ! n ! ( m + n + 1 ) ! \int_0^1 x^m (1-x)^n \, dx = \frac{m! \ n!}{(m+n+1)!}

I am not familiar with beta function right now. So, I did it by substitution.

Aritra Kundu - 5 years, 10 months ago

Beta Functions?

Jun Arro Estrella - 5 years, 10 months ago

And both substn and beta fn require calculation

Swapnil Vatsal - 3 years, 11 months ago

Use Wallis Formula

Kailash Agrawal - 3 years, 9 months ago

Trigonometry substitution is lengthy

Roy Satyam - 3 years, 6 months ago

At least I know how to do this by trig substitution rather than some fancy beta function

Yinchen Wu - 3 years, 5 months ago

You can use the trigonometric representation of the beta function. That makes it really easy.

N. Aadhaar Murty - 9 months, 3 weeks ago
Laurent Shorts
Mar 31, 2016

We have: ∫ 0 1 x n ( 1 − x 2 ) m d x {\displaystyle \int_{0}^{1}}x^{n}(1-x^{2})^{m}\mathrm{d}x = ( x n + 1 n + 1 ( 1 − x 2 ) m ) ∣ 0 1 ⏟ = 0 − ∫ 0 1 x n + 1 n + 1 m ( 1 − x 2 ) m − 1 ( − 2 x ) d x \,\,\,=\,\,\,\underset{=0}{\underbrace{\biggl(\dfrac{x^{n+1}}{n+1}(1-x^{2})^{m}\biggr)\biggl|_{0}^{1}}}\,\,\,-\,\,\,{\displaystyle \int_{0}^{1}}\dfrac{x^{n+1}}{n+1}m(1-x^{2})^{m-1}(-2x)\mathrm{d}x = 2 m n + 1 ∫ 0 1 x n + 1 ( 1 − x 2 ) m − 1 d x ={\displaystyle \dfrac{2m}{n+1}\int_{0}^{1}}x^{n+1}(1-x^{2})^{m-1}\mathrm{d}x

Therefore:

∫ 0 1 x 9 ( 1 − x 2 ) 9 d x = 18 10 ⋅ 16 12 ⋅ 14 14 ⋅ … ⋅ 2 26 ⋅ ∫ 0 1 x 27 d x {\displaystyle \int_{0}^{1}}x^{9}(1-x^{2})^{9}\mathrm{d}x\,\,\,=\,\,\,\dfrac{18}{10}·\dfrac{16}{12}·\dfrac{14}{14}·\ldots·\dfrac{2}{26}\,\,·\,\,{\displaystyle \int_{0}^{1}}x^{27}\mathrm{d}x = 9 ⋅ 8 ⋅ … ⋅ 1 5 ⋅ 6 ⋅ … ⋅ 13 ⋅ 1 28 = 4 ⋅ 3 ⋅ 2 ⋅ 1 10 ⋅ 11 ⋅ 12 ⋅ 13 ⋅ 28 = 1 10 ⋅ 11 ⋅ 13 ⋅ 14 =\dfrac{9·8·\ldots·1}{5·6·\ldots·13}\,·\,\dfrac{1}{28}=\dfrac{4·3·2·1}{10·11·12·13\,·\,28}=\dfrac{1}{10·11·13·14} = 1 2 0 ′ 020 =\dfrac{1}{20'020}

19 Helpful 0 Interesting 0 Brilliant 0 Confused

I particularly started using Gamma from today, so ∫ 0 π / 2 cos ⁡ m x sin ⁡ n x d x = Γ ( m + 1 2 ) Γ ( n + 1 2 ) 2 Γ ( m + n + 2 2 ) \int_0^{\pi/2} \cos^m x \sin^n x\, dx = \dfrac{\Gamma\left(\dfrac{m+1}2\right)\Gamma\left(\dfrac{n+1}2\right)}{2\Gamma\left(\dfrac{m+n+2}2\right)}

So, substituting x = sin ⁡ θ d x = cos ⁡ θ d θ I = ∫ 0 π / 2 cos ⁡ 19 θ sin ⁡ 9 θ d θ = Γ ( 20 2 ) Γ ( 10 2 ) 2 Γ ( 30 2 ) I = 9 ! ⋅ 4 ! 2 ⋅ 14 ! = 1 20020 x = \sin\theta\\dx = \cos \theta ~d\theta\\\ I = \int_0^{\pi/2} \cos^{19} \theta \sin^9 \theta\, d\theta = \dfrac{\Gamma\left(\dfrac{20}2\right)\Gamma\left(\dfrac{10}2\right)}{2\Gamma\left(\dfrac{30}2\right)}\\I = \dfrac{9!\cdot 4!}{2\cdot 14!} = \dfrac1{20020}

Or as an alternative, Beta also helps. Substituting x 2 = t 2 x d x = d t I = ∫ 0 1 ( 1 − t ) 9 t 4 d t = B ( 10 , 5 ) x^2 = t\\2x~dx = dt\\I = \int_0^1 (1-t)^9 t^4~dt =\mathrm B(10, 5)

And then going back to Gamma, B ( x , y ) = Γ ( x ) Γ ( y ) Γ ( x + y ) \mathrm B(x,y) = \dfrac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}

So, I = B ( 10 , 5 ) = Γ ( 10 ) Γ ( 5 ) Γ ( 15 ) = 9 ! ⋅ 4 ! 2 ⋅ 14 ! = 1 20020 I =\mathrm B(10, 5) = \dfrac{\Gamma(10)\Gamma(5)}{\Gamma(15)} = \dfrac{9!\cdot 4!}{2\cdot 14!} = \dfrac1{20020}

Note some important properties: Γ ( x ) = ∫ 0 ∞ e − t t x − 1 d t B ( x , y ) = B ( y , x ) = ∫ 0 1 t x − 1 ( 1 − t ) y − 1 d t ∫ 0 π / 2 cos ⁡ m x sin ⁡ n x d x = Γ ( m + 1 2 ) Γ ( n + 1 2 ) 2 Γ ( m + n + 2 2 ) B ( x , y ) = Γ ( x ) Γ ( y ) Γ ( x + y ) Γ ( 1 2 ) = π \Gamma(x) = \int_0^\infty e^{-t} t^{x-1} dt\\\\\mathrm B(x,y) =\mathrm B(y,x) = \int_0^1 t^{x-1}(1-t)^{y-1} dt\\\\ \int_0^{\pi/2} \cos^m x \sin^n x\, dx = \dfrac{\Gamma\left(\dfrac{m+1}2\right)\Gamma\left(\dfrac{n+1}2\right)}{2\Gamma\left(\dfrac{m+n+2}2\right)}\\\\\mathrm B(x,y) = \dfrac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}\\\\\Gamma\left(\dfrac 12\right) = \sqrt \pi

6 Helpful 0 Interesting 0 Brilliant 0 Confused
Shivam Mishra
Feb 24, 2016

'beta' makes it a child's play!!!!

0 Helpful 0 Interesting 1 Brilliant 0 Confused

We can directly use Wallis Formula by just plugging in x=Sinz

Kailash Agrawal - 3 years, 9 months ago

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