Question Number 101319 by mhmd last updated on 01/Jul/20

find the area bounded the parabola   y=4x^2    and  y=8−4x^2   ?  by using intigral?  help me

Commented bysmridha last updated on 01/Jul/20

the area under the curves(here parabola)  A=2[∫_0 ^1 (8−4x^2 )dx−∫_0 ^1 4x^2 dx]      =2[8x−(8/3)x^3 ]_0 ^1 =2[(8/1)−(8/3)]=((32)/3).

Commented bymhmd last updated on 01/Jul/20

very thank sir

Commented bysmridha last updated on 01/Jul/20

welcome...I think you can draw the picture..

Commented bymhmd last updated on 01/Jul/20

yes sir

Answered by bobhans last updated on 01/Jul/20

intercept : 4x^2  = 8−4x^2  ; 8x^2  = 8 ; x = ± 1  test x = (1/2) → { ((y=4×(1/4)=1)),((y=8−4×(1/4)= 7)) :}  8−4x^2  ≥ 4x^2  , so the area =   2∫_0 ^1  (8−8x^2 )dx = 2[(8x−(8/3)x^3 )]_0 ^1   = 2(8−(8/3)) = 2×((16)/3) = ((32)/3) ★

Commented bybramlex last updated on 02/Jul/20

 □♠★

Commented bymhmd last updated on 02/Jul/20

wow ! very nice sir thank you