MCQ
Let $n$ be a positive integer. For a real number $x$, let $[x]$ denote the largest integer not exceeding $x$ and $\{x\}=x-[x]$. Then, $\int \limits_1^{n+1} \frac{(\{x\})^{[x]}}{[x]} d x$ is equal to
  • A
    $\log _e(n)$
  • B
    $\frac{1}{n+1}$
  • $\frac{n}{n+1}$
  • D
    $1+\frac{1}{2}+\ldots . .+\frac{1}{n}$

Answer

Correct option: C.
$\frac{n}{n+1}$
c
(c)

Let

$I=\int \limits_1^{n+1} \frac{(\{x\})^{[x]}}{[x]} d x$

$I=\int \limits_1^2 \frac{\{x\}}{1} d x+\int \limits_2^3 \frac{\{x\}^2}{2} d x+\int \limits_3^4 \frac{\{x\}^2}{3} d x+$ $\cdots \int \limits_n^{n+1} \frac{\{x\}^n}{n} d x$

$\Rightarrow \quad I=\int \limits_0^1\left(\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+\ldots+\frac{x^n}{n}\right) d x$

$\Rightarrow I=\left[\frac{x^2}{1 \times 2}+\frac{x^3}{2 \times 3}+\frac{x^4}{3 \times 4}+\ldots+\frac{x^{n+1}}{n(n+1)}\right]_0^1$

$\Rightarrow I=\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+\ldots+\frac{1}{n(n+1)}$

$\Rightarrow I=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots$

$\Rightarrow I=1-\frac{1}{n+1}=\frac{n+1-1}{n+1}=\frac{n}{n+1}$

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