can not figure out problem.

Question

(the area of the region in the first quadrant between the graph of y=x sqrt (4-x^2) and the axis is ?

a(2/3)sqrt2
b 8/3
c 2sqrt2
d 2sqrt3
e 16/3

Answer 1

f(x)>0 on(0,2) and f(0)=f(2)=0
The primitive of f(x) is
-(4-x^2)sqrt(4-x^2)/3
The area=-(4-2^2)sqrt(4-2^2)/3+(4-0^2)sqrt(4-0^)/3=8/3

Answer 2

I agree with Mgraph.