badge image

Enroll For Free Now & Improve Your Performance.

×
User Icon
User Icon
User Icon
User Icon
User Icon

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
Menu
Grade:

                        

Please solve the question in the image attached Answer of this question is Option 2

one year ago

Answers : (1)

Aditya Gupta
2069 Points
							
hello manjula, this ques can be easily solved  by drawing a graph of the 2 functions. further note that [x+1] can be written as [x] + 1 where [.] represents gif. (this follows from std properties of gif)
once you draw the graph, you ll notice that y= x^2 and y= [x] + 1 tend to meet at x=1. also, y= [x] + 1 is greater than x^2 when x is less than 1 but greater than 0.
so area bounded A= ∫ from 0 to 1 ([x] + 1 – x^2) dx
note that in x belonging to (0, 1), [x]= 0
so A= ∫ from 0 to 1 (1 – x^2) dx
= x – x^3/3 from 0 to 1
= 1 – 1/3
2/3
kindly approve :)
one year ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

  • 731 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution


Course Features

  • 51 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details