SimPly InteGrate It !!!

Calculus Level 5

If $\alpha$ = $\displaystyle \int_{0}^{1}(\prod _{r=1}^{2015}(x+r))\cdot (\sum_{k=1}^{2015}(\dfrac{1}{x+k})) \mathrm dx$

Then $\alpha$ can be expressed as $A!B$ .

Find $\dfrac{A+B}{25} + \dfrac{9,573}{A+B+\dfrac{A}{5}-212}$ .

$\bullet \textbf{Details}$

Answer Upto 3 Decimal Places

$A,B$ Both Are $\textbf{4 Digits Numbers}$

2 solutions

Incredible Mind
Feb 10, 2015

easy indefinite integral..Got it just by looking

Kartik Sharma
Feb 7, 2015

Well, $\int_{0}^{1}{(x + 1)(x+2)(x+3)...(x+2015)(\frac{1}{x+1} + \frac{1}{x+2} + .... +\frac{1}{x+2015})}dx$

First solution coming in anybody's mind would be IBP.

So, it is so.

$dv = \sum_{k=1}^{2015}{\frac{1}{x+k}} dx$

$v = ln(x+1) + ln(x+2) + ln(x+3) + .... ln(x+2015)$

$u = \prod_{r=1}^{2015}{x+r}$

$u = {e}^{ln(x + 1) + ln(x+2) + .... + ln(x+2015)}$

$du = (x+1)(x+2)(x+3)...(x+2015)(\frac{1}{x+1} + \frac{1}{x+2} + ....\frac{1}{x+2015}) dx$

But wait, this is the integral we started with. Therefore, $u = \alpha = \frac{(x+2015)!}{x!}$

Substituting,

$\frac{2016!}{1!} - \frac{2015!}{0!} = 2015.2015!$

Now, all we need to do is substitute values.

Vraj Mehta But $2015.2015!$ is not the only way you can write as $A!B$ . There can be many like $\frac{2015}{2016}. 2016!$ , $4060225.2014!$ etc.

I Hope That Discrepancy is removed Now

Vraj Mehta - 6 years, 4 months ago

Hmm Yeah that's fine now. BTW, why such a random answer in decimals(just asking)?

Kartik Sharma - 6 years, 4 months ago