Click to Chat

1800-2000-838

+91-120-4616500

CART 0

• 0

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
        if[x] denotes the greatest integer less than or equal to x then integral of [1/1+x2]dx in limits from 1 to infinity =
7 years ago

Jitender Singh
IIT Delhi
158 Points
										Ans: 1Sol:$I = \int_{1}^{\infty }[\frac{1}{1+x^{2}}]dx$$I = \int_{1}^{2}[\frac{1}{1+x^{2}}]dx + \int_{2}^{3}[\frac{1}{1+x^{2}}]dx + .....$$1\leq x\leq 2\Rightarrow [\frac{1}{1+x^{2}}] = 1$$2\leq x\leq 3\Rightarrow [\frac{1}{1+x^{2}}] = 0$Similarly zero for the next intervals.$I = \int_{1}^{2}dx = 1$Cheers!Thanks & RegardsJitender SinghIIT DelhiaskIITians Faculty

3 years ago
Jitender Singh
13 Points
										Ans: 0There is one minor mistake in the above solution. Sorry about that. When x varies from 1 to 2, then integral part of the 1/(1+x2) should be equal to zero. So the integral value should be zero. Thanks & RegardsJitender SinghIIT DelhiaskIITians Faculty

3 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Discuss with Askiitians Tutors

View all Questions »
• Complete AIPMT/AIIMS Course and Test Series
• OFFERED PRICE: Rs. 15,900
• View Details
• Complete JEE Main/Advanced Course and Test Series
• OFFERED PRICE: Rs. 15,900
• View Details