Join now for JEE/NEET and also prepare for Boards Join now for JEE/NEET and also prepare for Boards. Register Now
Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
The area enclosed by the curve y 2 + x 4 = x 2 is- (a)2/3 (b)4/3 (c)8/3 (d)10/3 The area enclosed by the curve y2 + x4 = x2 is-(a)2/3(b)4/3(c)8/3(d)10/3
Dear student This curve is symmetrical along the x and y axis, so we can integrate the area in the first quadrant and multiply it by 4. y = x(√(1-x^2)) A = ∫ 0->1 x(√(1-x^2)) dx A = -1/3 (1-x^2)^(3/2) | 0->1 A = 1/3 Total Area is 1/3 x 4 = 4/3
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -