Download App

STD 12 - 7.2 definite integral — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 7.2 definite integral1 Mark
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

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

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

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

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

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

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

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

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from STD 12 - 7.2 definite integral→STD 12 - 7.2 definite integral — all question types→Mathematics STD 12 Science question papers→CBSE Board STD 12 Science question papers→CBSE Board question paper generator→

Similar questions

There are two value of a which makes the determinant $\triangle=\begin{vmatrix}1&-2&5\\2&\text{a}&-1\\0&4&2\text{a}\end{vmatrix}$ equal to $86$. The sum of these two values is:
→
Choose the correct answer from the given four options. You are given that $A$ and $B$ are two events such that $\text{P}(\text{B})=\frac{3}{5},\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{2}$ and $\text{P}(\text{A}\cup\text{B})=\frac{4}{5},$ then $P(A)$ equals :
→
Value of the definite integral   $\int\limits_{ - 1/2}^{1/2}$$( sin^{-1}(3x- 4x^3)- cos^{-1}(4x^3- 3x) )dx$
→
Let $f : R \to R$ , where $f(x) = \frac{{a{x^2} + ax + b}}{{ax + b}}$ , then
→
Solution of the differential equation $xdy = (y + xy^3 (1 + log_ex))\ dx$ is (Where $C$ is arbitary constant)
→
$\int_{}^{} {\frac{1}{x}{{\sec }^2}(\log x)dx = } $
→
The value of $\big(\vec{\text{a}}\times\vec{\text{b}}\big)^2$ is:
→
Let $g$ is the inverse function of $f \,\&\, f ' (x) = \frac{{{x^{10}}}}{{\left( {1\,\, + \,\,{x^2}} \right)}}$. If $g(2) = a$ then $g ' (2)$ is equal to
→
The value of $c$ in Rolle's theorem for the function $\text{f}(\text{x})=\frac{\text{x}(\text{x}+1)}{\text{e}^{\text{x}}}$ defined on $[-1, 0]$ is:
→
Let a line $l$ pass through the origin and be perpendicular to the lines

$l_1: \overrightarrow{ r }=(\hat{ i }-11 \hat{ j }-7 \hat{ k })+\lambda(\hat{ i }+2 \hat{ j }+3 \hat{ k }), \lambda \in R$

and $l_2: \overrightarrow{ r }=(-\hat{ i }+\hat{ k })+\mu(2 \hat{ i }+2 \hat{ j }+\hat{ k }), \mu \in R$.

If $P$ is the point of intersection of $l$ and $l_1$, and $Q (\alpha$ $, \beta, \gamma)$ is the foot of perpendicular from $P$ on $l_2$, then $9(\alpha+\beta+\gamma)$ is equal to $..........$.

→