What is the area bounded by y= 16-x^2 , y 0 and the x -axis?

Step-by-step solution

aligned & y= 16-x^2 & As, y 0,16-x^2 0 & x [-4,4] aligned Now, from I= a^2-x^2 d x= x a^2-x^2 2 + a^2 2 ⁻¹ ( x a )+C Here a^2=16 Area bounded by the curve aligned I & = _ -4 ^4 (4)^2-x^2 d x & = [ x 16-x^2 2 ]_ -4 ^4+ 16 2 [ ⁻¹ ( x 4 ) ]_ -4 ^4 aligned aligned & =0+8 [ ⁻¹ ( 4 4 )- ⁻¹ (- 4 4 ) ] & =8 [ ⁻¹(1)+ ⁻¹(1) ] & =16 ⁻¹(1)=16 2 =8 square units aligned