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  • Need clarity, kindly explain! - Integral Calculus - JEE Main-6

\lim_{n\rightarrow \infty }\frac{1^{p}+2^{p}+3^{p}+...+n^{p}}{n^{p+1}}\; \; is

  • Option 1)

    \frac{1}{P+1}\;

  • Option 2)

    \; \frac{1}{1-P}\;

  • Option 3)

    \; \frac{1}{P}-\frac{1}{P-1}\;

  • Option 4)

    \; \frac{1}{P+2}

 

Answers (1)

best_answer

As we learnt in 

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

 

 \Rightarrow \lim_{n\rightarrow \infty} \frac{1^{p}+2^{p}+3^{p}+.........+n^{p}}{n^{p+1}}

\Rightarrow \lim_{n\rightarrow \infty} \sum \frac{r^{p}}{n^{p+1}}

\Rightarrow \lim_{n\rightarrow \infty} \frac{1}{n}\sum (\frac{r}{n})^{p}

\Rightarrow \int_{0}^{1}x^{p}dx=\left [ \frac{x^{p+1}}{p+1} \right ]_{0}^{1}=\frac{1}{p+1}


Option 1)

\frac{1}{P+1}\;

This solution is correct 

Option 2)

\; \frac{1}{1-P}\;

This solution is incorrect 

Option 3)

\; \frac{1}{P}-\frac{1}{P-1}\;

This solution is incorrect 

Option 4)

\; \frac{1}{P+2}

This solution is incorrect 

Posted by

Plabita

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