Integral of a defined function-

Calculus Level 5

Let a function f ( x ) f(x) be defined as:

f ( x ) = ∑ r = 0 x r 2 ( x r ) . f(x) = \displaystyle\sum_{r=0}^x r^{2}~\dbinom{x}{r}.

Find the value of

24 ∫ 0 1 f ( x ) 2 x d x . 24 \displaystyle\int_0^{1} \dfrac{f(x)}{2^{x}} dx.


The answer is 5.

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Avineil Jain by Avineil Jain , 24, Netherlands

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2 solutions

Avineil Jain
Jun 14, 2014

f ( x ) = ∑ r = 0 x r 2 ( x r ) f(x) = \displaystyle\sum_{r=0}^x r^{2}~\dbinom{x}{r}

f ( x ) = ∑ r = 0 x ( r ( r − 1 ) + r ) ( x r ) f(x) = \displaystyle\sum_{r=0}^x (r(r-1) + r) ~\dbinom{x}{r}

f ( x ) = ∑ r = 0 x r ( r − 1 ) ( x r ) + ∑ r = 0 x r ( x r ) f(x) = \displaystyle\sum_{r=0}^x r(r-1)\dbinom{x}{r} + \displaystyle\sum_{r=0}^x r\dbinom{x}{r}

We now use the famous identity -

( n r ) = n r ( n − 1 r − 1 ) \dbinom{n}{r} = \dfrac{n}{r}~\dbinom{n-1}{r-1}

f ( x ) = ∑ r = 2 x x ( x − 1 ) ( x − 2 r − 2 ) + ∑ r = 1 x x ( x − 1 r − 1 ) f(x) = \displaystyle\sum_{r=2}^x x(x-1) \dbinom{x-2}{r-2} + \displaystyle\sum_{r=1}^x x\dbinom{x-1}{r-1}

f ( x ) = x ( x − 1 ) 2 x − 2 + x 2 x − 1 f(x) = x(x-1)~2^{x-2} + x~2^{x-1}

f ( x ) = x ( x + 1 ) 2 x − 2 f(x) = x(x+1)~2^{x-2}

24 ∫ 0 1 f ( x ) 2 x = 24 ∫ 0 1 x ( x + 1 ) 4 24 \displaystyle\int_{0}^{1} \dfrac{f(x)}{2^{x}} = 24 \displaystyle\int_{0}^{1} \dfrac{x(x+1)}{4}

24 ∫ 0 1 f ( x ) 2 x = 5 24 \displaystyle\int_{0}^{1} \dfrac{f(x)}{2^{x}} = 5

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@Avineil Jain @Calvin Lin Is the Binomial Coefficient defined for non-integral values of x x ?

We generally use Γ ( 1 + x ) = ∫ 0 ∞ t x e − t d t \Gamma(1+x) =\displaystyle\int_{0}^{\infty} t^x e^{-t}\mathrm{d}t instead of x ! x! for non-integral values of x x , don't we?

Pratik Shastri - 6 years, 8 months ago
Pradeep Ch
Jun 14, 2014

( 1 + t ) x = ( x 0 ) + ( x 1 ) t + ( x 2 ) t 2 + ( x 3 ) t 3 + . . . . ( x x ) t x d i f f e r e n t i a t e i t w . r . t t . x ( 1 + t ) x − 1 = ( x 1 ) + 2 ( x 2 ) t + 3 ( x 3 ) t 2 + . . . . x ( x x ) t x − 1 m u l t i p l y L H S a n d R H S b y t . x t ( 1 + t ) x = ( x 1 ) t + 2 ( x 2 ) t 2 + 3 ( x 3 ) t 3 + . . . . x ( x x ) t x a g a i n d i f f e r e n t i a t e w . r . t t . x [ ( 1 + t ) x − 1 + t ( x − 1 ) ( 1 + t ) x − 2 ] = ( x 1 ) + 2 2 ( x 2 ) t + 3 2 ( x 3 ) t 2 + . . . . x 2 ( x x ) t x − 1 p u t t = 1 x [ ( 2 ) x − 1 + ( x − 1 ) ( 2 ) x − 2 ] = f ( x ) s o , f ( x ) 2 x = x [ 1 2 + ( x − 1 ) 1 4 ] = x ( x + 1 ) 4 s o , 24 ∫ 0 1 f ( x ) 2 x = 24 ∫ 0 1 x ( x + 1 ) 4 = 5 { (1+t) }^{ x }\quad =\quad \left( \begin{matrix} x \\ 0 \end{matrix} \right) \quad +\quad \left( \begin{matrix} x \\ 1 \end{matrix} \right) t\quad +\quad \left( \begin{matrix} x \\ 2 \end{matrix} \right) { t }^{ 2 }\quad +\quad \left( \begin{matrix} x \\ 3 \end{matrix} \right) { t }^{ 3 }\quad +\quad ....\quad \left( \begin{matrix} x \\ x \end{matrix} \right) { t }^{ x }\\ differentiate\quad it\quad w.r.t\quad t.\\ x{ (1+t) }^{ x-1 }\quad =\quad \left( \begin{matrix} x \\ 1 \end{matrix} \right) \quad +\quad 2\left( \begin{matrix} x \\ 2 \end{matrix} \right) t\quad +\quad 3\left( \begin{matrix} x \\ 3 \end{matrix} \right) { t }^{ 2 }\quad +\quad ....\quad x\left( \begin{matrix} x \\ x \end{matrix} \right) { t }^{ x-1 }\\ multiply\quad LHS\quad and\quad RHS\quad by\quad t.\\ xt{ (1+t) }^{ x }\quad =\quad \left( \begin{matrix} x \\ 1 \end{matrix} \right) t\quad +\quad 2\left( \begin{matrix} x \\ 2 \end{matrix} \right) { t }^{ 2 }\quad +\quad 3\left( \begin{matrix} x \\ 3 \end{matrix} \right) { t }^{ 3 }\quad +\quad ....\quad x\left( \begin{matrix} x \\ x \end{matrix} \right) { t }^{ x }\\ again\quad differentiate\quad w.r.t\quad t.\\ x[{ (1+t) }^{ x-1 }\quad +\quad t(x-1){ (1+t) }^{ x-2 }]\quad =\quad \quad \left( \begin{matrix} x \\ 1 \end{matrix} \right) \quad +\quad { 2 }^{ 2 }\left( \begin{matrix} x \\ 2 \end{matrix} \right) t\quad +\quad { 3 }^{ 2 }\left( \begin{matrix} x \\ 3 \end{matrix} \right) { t }^{ 2 }\quad +\quad ....\quad { x }^{ 2 }\left( \begin{matrix} x \\ x \end{matrix} \right) { t }^{ x-1 }\\ put\quad t=1\\ x[{ (2) }^{ x-1 }\quad +\quad (x-1){ (2) }^{ x-2 }]\quad =\quad f(x)\\ \\ so,\quad \frac { f(x) }{ { 2 }^{ x } } \quad =\quad x\left[ \frac { 1 }{ 2 } \quad +\quad (x-1)\frac { 1 }{ 4 } \right] \quad =\quad \frac { x(x+1) }{ 4 } \\ so,\quad 24\int _{ 0 }^{ 1 }{ \frac { f(x) }{ { 2 }^{ x } } } \quad =\quad 24\int _{ 0 }^{ 1 }{ \frac { x(x+1) }{ 4 } } =\quad 5

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