Integral Calculus> options 3/π log2 2/πlog2 3/π log3 3/π...

options\t3/π log2\t2/πlog2\t3/π log3\t3/π

Aditya Kartikeya , 10 Years ago

Grade 10

1 Answers

Jitender Singh

Last Activity: 10 Years ago

Ans:
Hello Student,
Please find answer to your question below

$L = \lim_{n\rightarrow \infty }[tan\frac{\pi }{3n}+tan\frac{2\pi }{3n}+tan\frac{3\pi }{3n}+.................+tan\frac{\pi }{3}]\frac{1}{n}$
$L = \lim_{n\rightarrow \infty }[\sum_{r = 1}^{n}tan\frac{r\pi }{3n}]\frac{1}{n}$
$L = \lim_{n\rightarrow \infty }[\sum_{r = 1}^{n}\frac{1}{n}.tan\frac{r\pi }{3n}]$
Now we are going to convert it into definite integral using the definition
$\frac{r}{n} = x,\sum_{r = 1}^{n}\frac{1}{n} = dx$
Limit is from 0 to 1.
$L = \int_{0}^{1}tan\frac{\pi x}{3}dx$
$L = \int_{0}^{1}\frac{sin\frac{\pi x}{3}}{cos\frac{\pi x}{3}}dx$
$L = \frac{3}{\pi }\int_{0}^{1}\frac{\frac{\pi }{3}.sin\frac{\pi x}{3}}{cos\frac{\pi x}{3}}dx$
$cos\frac{\pi x}{3} = t$
$-\frac{\pi }{3}sin\frac{\pi x}{3}dx = dt$
$x = 0\rightarrow t = 1$
$x = 1\rightarrow t = \frac{1}{2}$
$L = \frac{3}{\pi }\int_{1}^{1/2}\frac{-1}{t}dt$
$L = \frac{3}{\pi }\int_{1/2}^{1}\frac{1}{t}dt$
$L = \frac{3}{\pi }[logt]_{1/2}^{1}$
$L = \frac{3}{\pi }[-log(1/2)]$
$L = \frac{3}{\pi }[log(2)]$
Option (1) is correct.