MY CART (5)

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

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: R

There are no items in this cart.
Continue Shopping
Menu
Get instant 20% OFF on Online Material.
coupon code: MOB20 | View Course list

Get extra R 480 off
USE CODE: asknikunj

				   

Area bouned by the region [x]^2=[y]^2 of x belongs to [1,5] (where [.] denotes the greatest integer function)

4 years ago

Share

Answers : (1)

										

Hi Debadutta,


 


Let 1≤x<2. So [x] = 1.


And [y]^2 = 1 or [y] = ± 1.


means 1≤y<2 or -1≤y<0.


So in this region you will have two squares each of area 1 unit.


So when 1≤x<2 the area will be 2 sq units.


 


Similarly when 2≤x<3 the area = 2 sq untis


And 3≤x<4, area = 2 sq untis.


and 4≤x<5, area = 2 sq units.


 


So total area in the region would be 2+2+2+2 = 8 sq untis.


 


Regards,


Ashwin (IIT Madras).

4 years ago

Post Your Answer

Other Related Questions on Integral Calculus

Please it would be grateful if anybody could tell me the integration of xtan(x)dx
 
 
As it will result in the recursive equation and therefore it will be requried to convert into some special function.
  img
Vijay Mukati one year ago
 
I will try to provide it whenever I will be free. Thanks.
  img
Vijay Mukati one year ago
 
HIint. Please use the product rule for the integration here. Consider the x as 1 st term and tanx as second. Thanks.
  img
Vijay Mukati one year ago
is arihant differential and inegral calculus by amit agrwal is good book for iit and board exam
 
 
Yes it is a good book indeed but it is of advanced type so it may not be use for board exam.
 
VINEETH N one year ago
plz send the full solution of this question
 
 
 
jagdish singh singh 9 months ago
Show that ln(1+x)>x for all x>0.
 
 
question printed is incorrect.Correct question:- Show that ln(1+x)0.
 
subham mohanty one year ago
Sir can you tell me the solution of this complex number question in the attachment? The answer is A,D
 
 
Several ways to approach this: Write where So since Hence So either z is purely imaginary or z=0. (z=0 is possible when z 1 =-z 2 which is permitted with the restrictions mentioned in the...
 
mycroft holmes 2 days ago
 
put z1 = x1 +iy1 and'z2 = x2 + iy2 simple and also thiss is given as: (x1)^2 + (y1)^2 = (x2)^2 + (y2)^2. Us ethis info.
 
Vikas TU 3 days ago
Find d (x 4 +x 2 +1) / x 2 +x+1 dx Please tell me how do this question.
 
 
on dividing numerator with denominator:(x^4+x^2+1)/(x^2+x+1)= =((x^2+x+1)(x^2-x+1)+1)/(x^2+x+1) =(x^2-x+1)+1/(x^2+x+1) on differentiating the resulted equation we will get...
 
SREEKANTH 2 months ago
 
Numerator is divisible by denominator . On dividing numerator by denominator the it is reduced to x 2 -x +1 differentiating this we get 2x-1 as answer
 
Ajay 2 months ago
View all Questions »

  • Complete JEE Main/Advanced Course and Test Series
  • OFFERED PRICE: R 15,000
  • View Details
Get extra R 4,500 off
USE CODE: asknikunj

Get extra R 480 off
USE CODE: asknikunj

More Questions On Integral Calculus

Ask Experts

Have any Question? Ask Experts

Post Question

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