The value of int_{0}^{1} left( prod_{r=1}^{n} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $f(x) = \prod_{r=1}^{n} (x+r) = (x+1)(x+2)\dots(x+n)$.Using the logarithmic differentiation rule,$\frac{d}{dx} \ln(f(x)) = \frac{f'(x)}{f(x)}

Vedclass · Accepted Answer

$n \cdot n!$

Vedclass · Answer

(D) Let $f(x) = \prod_{r=1}^{n} (x+r) = (x+1)(x+2)\dots(x+n)$.Using the logarithmic differentiation rule,$\frac{d}{dx} \ln(f(x)) = \frac{f'(x)}{f(x)} = \sum_{k=1}^{n} \frac{1}{x+k}$.Thus,$f'(x) = \left( \prod_{r=1}^{n} (x+r) \right) \left( \sum_{k=1}^{n} \frac{1}{x+k} \right)$.The integral becomes $\int_{0}^{1} f'(x) dx = [f(x)]_{0}^{1} = f(1) - f(0)$.$f(1) = (1+1)(1+2)\dots(1+n) = 2 \cdot 3 \dots (n+1) = (n+1)!$.$f(0) = (0+1)(0+2)\dots(0+n) = 1 \cdot 2 \dots n = n!$.Therefore,the integral value is $(n+1)! - n! = n!(n+1-1) = n \cdot n!$.

The value of $\int_{0}^{1} \left( \prod_{r=1}^{n} (x+r) \right) \left( \sum_{k=1}^{n} \frac{1}{x+k} \right) dx$ equals

Similar Questions

Reserpine is obtained from ...... .

If $s$ and $p$ are respectively the sum and the product of the slopes of the lines $3x^2 - 2xy - 15y^2 = 0$,then $s:p$ is equal to

If $X$ is a Poisson variate such that $P(X=1)=P(X=2)$,then $P(X=4)$ is equal to

The equation of the tangent at $(-4, -4)$ on the curve $x^2 = -4y$ is

This is the only type of pollination which during pollination brings genetically different types of pollen grains to the stigma.

Vedclass Test Series

Exam Paper Generator

Online Exam Module