(Product)*(summation) under integral?

Calculus Level 4

Solve: ∫ 0 1 ( ∏ r = 1 n ( x + r ) ) ∑ k = 1 n 1 x + k d x \large\displaystyle\int _{ 0 }^{ 1 }{ \left( \prod _{ r=1 }^{ n }{ (x+r) } \right) \sum _{ k=1 }^{ n }{ \frac { 1 }{ x+k } \text{d}x } }

n! n (n+1)! n.n!

Kunal Gupta by Kunal Gupta , 23, India

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1 solution

Kunal Gupta
Aug 4, 2014

This Problem comes with a little observation: = ∫ 0 1 ( ( x + 1 ) ( x + 2 ) ( x + 3 ) . . . . ( x + n ) ) ( 1 x + 1 + 1 x + 2 + . . . . . . . + 1 x + n ) w h i c h b e c o m e s ; = ∫ 0 1 d ( ( x + 1 ) ( x + 2 ) ( x + 3 ) . . . . . . ( x + n ) ) = ⌈ ( x + 1 ) ( x + 2 ) ( x + 3 ) . . . . ( x + n ) ⌉ 0 1 = ( n + 1 ) ! − ( n ) ! = n ! ( n + 1 − 1 ) = n . n ! ( a n s w e r ) =\displaystyle\quad \int _{ 0 }^{ 1 }{ ((x+1)(x+2)(x+3)....(x+n) } )\left( \frac { 1 }{ x+1 } +\frac { 1 }{ x+2 } +.......+\frac { 1 }{ x+n } \right) \\ which\quad becomes;\\ =\displaystyle \int _{ 0 }^{ 1 }{ d\left( (x+1)(x+2)(x+3)......(x+n) \right) \quad =\quad \left\lceil (x+1)(x+2)(x+3)....(x+n) \right\rceil } _{ 0 }^{ 1 }\\ =(n+1)!-(n)!\\ =n!(n+1-1)=\quad \quad \quad \large \boxed { n.n! } \quad (answer)

13 Helpful 0 Interesting 1 Brilliant 0 Confused

We wouldn't memorize that thing for sure that it is the derivative of that.

Rather, try using IBP and then one will surely take u = ( x + 1 ) ( x + 2 ) ( x + 3 ) . . . ( x + n ) u = (x+1)(x+2)(x+3)...(x+n) and then while taking derivative of this, one can realize better I guess.

Kartik Sharma - 6 years, 5 months ago

I think from u=(x+1)(x+2)...(x+n), if you take log on both sides and differentiate, the result follows.

Raghu Raman Ravi - 3 years, 10 months ago

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