Add and Multiply within a Definite Integral!

Calculus Level 3

Find the value of : ∫ 0 1 ( ∏ r = 1 n ( x + r ) ) ( ∑ k = 1 n 1 x + k ) d x \displaystyle\int_0^1 \left ( \displaystyle\prod_{r=1}^n(x+r) \right) \left ( \displaystyle\sum_{k=1}^n \dfrac{1}{x+k} \right) dx

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

Rishu Jaar by Rishu Jaar , 21, India

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1 solution

Put t = ∏ r = 1 n ( x + r ) \displaystyle t=\prod_{r=1}^n (x+r) , this will give d t = ( ∏ r = 1 n ( x + r ) ) ( ∑ r = 1 n 1 x + r ) \displaystyle dt= \left(\prod_{r=1}^n (x+r)\right)\left(\sum_{r=1}^n \dfrac{1}{x+r}\right) .

So basically the entire expression is substituted giving I = ∫ d t = t + C I=\int dt =t+C . This on putting the limits after substituting back x x we get , I = ( n + 1 ) ! − n ! = n . n ! I=(n+1)!-n!=n.n! and that gives the answer.

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