Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

option s √3/3 √3 2 2√3

options
  1. √3/3
  2. √3
  3. 2
  4. 2√3

Question Image
Grade:10

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
6 years ago
Ans:
Hello Student,
Please find answer to your question below

L = \lim_{n\rightarrow \infty }\frac{\pi }{6n}[sec^{2}(\frac{\pi }{6n})+sec^{2}(\frac{2\pi }{6n})+....................+sec^{2}(\frac{n\pi }{6n})]
L = \lim_{n\rightarrow \infty }\frac{\pi }{6n}[\sum_{r = 1}^{n}sec^{2}(\frac{r\pi }{6n})]
Apply the definition of definite integral we have
L = \int_{0}^{1}\frac{\pi }{6}sec^{2}(\frac{\pi x}{6})dx
tan\frac{\pi x}{6} = t
x = 0\rightarrow t = 0
x = 1\rightarrow t = \frac{1}{\sqrt{3}}
\frac{\pi }{6}sec^{2}\frac{\pi x}{6}dx = dt
L = \int_{0}^{1/\sqrt{3}}dt
L =(t)_{0}^{1/\sqrt{3}}
L = \frac{1}{\sqrt{3}}

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free