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\lim_{n\rightarrow \infty }\left [ \frac{1}{n^{2}}sec^{2}\: \frac{1}{n^{2}}+\frac{2}{n^{2}}sec^{2}\: \frac{4}{n^{2}}+....+\frac{1}{n}sec^{2}\; 1 \right ]    equals

  • Option 1)

    \frac{1}{2}cosec\, 1

  • Option 2)

    \frac{1}{2}sec\, 1

  • Option 3)

    \frac{1}{2}tan\, 1

  • Option 4)

    tan\, 1

 

Answers (1)

As learnt in concept

Walli's Method -

Definite integral by first principle

\int_{a}^{b}f(x)dx= \left ( b-a \right )\lim_{n \to \infty }\frac{1}{n}\left [ f(a) +f(a+h)+f(a+2h)....\right ]

where

h=\frac{b-a}{n}

- wherein

 

 \lim_{n\rightarrow \infty }\sum_{r=1}^{n}\frac{r}{n^{2}}sec^{2}\frac{r^{2}}{n^{2}}; Here \frac{r}{n}=x \ and\ \frac{1}{n}------------->0

We get, \int_{0}^{1}xsec^{2}x^{2}dx

Put x^{2}=t

By substitution, we get,

\frac{1}{2}\int_{0}^{1}2x\:sec^{2}x^{2}dx=\frac{1}{2}\int_{0}^{1}sec^{2}tdt

= \frac{1}{2}(tant)_{0}^{1}=\frac{1}{2}tan1

 


Option 1)

\frac{1}{2}cosec\, 1

This is incorrect

Option 2)

\frac{1}{2}sec\, 1

This is incorrect

Option 3)

\frac{1}{2}tan\, 1

This is correct

Option 4)

tan\, 1

This is incorrect

Posted by

Vakul

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