_ n 1 n _ r = 1 ^ 2n r n^2 + r^2 equals - Vidyadip

MCQ

$\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{r = 1}^{2n} {\frac{r}{{\sqrt {{n^2} + {r^2}} }}} $ equals

A
$1 + \sqrt 5 $
✓
$ - 1 + \sqrt 5 $
C
$ - 1 + \sqrt 2 $
D
$1 + \sqrt 2 $

✓

Answer

Correct option: B.

$ - 1 + \sqrt 5 $

b
(b) $L = \mathop {\lim }\limits_{n \to \infty } \,\,\sum\limits_{r = 1}^{2n} {} \frac{1}{n}\,.\,\frac{{r/n}}{{\sqrt {1 + {{(r/n)}^2}} }}$

$= \int_0^2 {} \frac{x}{{\sqrt {1 + {x^2}} }}\,dx = \sqrt 5 - 1$.

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

The solution set of the inequation $3 x+5 y<7$ is

If $\vec a\, = \,\vec i - 2\hat j + 3\hat k,\,\,\,\vec b = 2\vec i + 3\hat j - \hat k$ and $\vec c = \lambda \vec i + \hat j + (2\lambda - 1\hat k)$ are coplanar vectors, then $\lambda $ is equal to

If $A$ is a square matrix such that $A ^2= A,$ then $(I-A)^3+ A$ is equal to:

$\int_{}^{} {\frac{{2x{{\tan }^{ - 1}}{x^2}}}{{1 + {x^4}}}} \;dx = $

$\int_{}^{} {\sqrt {1 - \sin 2x} \;} dx = ........,\;\;x \in (0,\;\pi /4)$

If ${\cot ^{ - 1}}\frac{n}{\pi } > \frac{\pi }{6},\,\,n \in N$ , then the maximum value of $n$ is

The angle between the straight lines $\frac{\text{x}+1}{2}=\frac{\text{y}-2}{5}=\frac{\text{z}+3}{4}$ and $\frac{\text{x}-1}{1}=\frac{\text{y}+2}{2}=\frac{\text{z}-3}{-3}$ is:

For any matrix $A , A =\left[\begin{array}{cc}\alpha & -2 \\ -2 & \alpha\end{array}\right],\left| A ^3\right|=125$ then value of $\alpha$ is :

The value of $\sin \left( {2{{\tan }^{ - 1}}\left( {\frac{1}{3}} \right)} \right) + \cos ({\tan ^{ - 1}}2\sqrt 2 ) = $

If $f(x)=x+1$, find $\frac{d}{d x}(f o f)(x)$.