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.

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

Total Price: Rs.

There are no items in this cart.
Continue Shopping
Grade: 12


Kindly show me the solution for the given sum. It is taken from the January session of the JEE main examinations.

6 months ago

Answers : (1)

Aditya Gupta
2065 Points
this is a rather easy problem.
we simply need to apply the property: ∫a to b g(x)dx = ∫a to b g(a+b – x)dx
now, let J= (1/(a+b)) * ∫a to b x(f(x) + f(x+1))dx.......(1)
so J= (1/(a+b)) * ∫a to b (a+b – x)(f(a+b – x) + f(a+b – x+1))dx........(2)
now, since f(a+b+1 – x)= f(x) for all x, replace x by x+1 to get f(a+b – x)= f(x+1)
so, (2) becomes J= (1/(a+b)) * ∫a to b (a+b – x)(f(x+1) + f(x))dx......(3)
add (1) and (3)
2J= (1/(a+b)) * ∫a to b (a+b)(f(x+1) + f(x))dx
or 2J= ∫a to b (f(x+1) + f(x))dx
or 2J= ∫a to b f(x)dx + ∫a to b f(x+1)dx
now, ∫a to b f(x+1)dx= ∫a to b f(a+b – x+1)dx= ∫a to b f(x)dx
so, 2J= ∫a to b f(x)dx + ∫a to b f(x)dx
or J= ∫a to b f(x)dx
let x= y+1
then J= ∫a – 1 to b – 1 f(y+1)dy
since y is a dummy variable, replace it back by x.
so J= ∫a – 1 to b – 1 f(x+1)dx
option (3)
6 months 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
To Win!!! Click Here for details