Calculus question help please?

Question

The area of the region in the first quadrant between the graph of y = x (4-x^2) and the x-axis is
NB there is a square root over 4-x^2

(A) 2/3
(B) 8/3
(C) 2
(D) 16/3

Answer 1

(B) 8/3


Consider the region drawn out by that curve in the first quadrant (go through values from 0, 1, 2, ...) Notice that it crosses the x-axis again at x=2, and all numbers after that, it is transformed to the imaginary plane (as you take the square root of a negative number).

So integrate y from 0 to 2:

∫ x√(4-x²) dx

You should see by reconition that the x on the outside of the square root is the derivative of what is inside the square root. Which is useful, because it allows us to apply this rule of integration:

∫ f(x)^n dx
=
{f(x)^(n+1)}/[(n+1)f'(x)] + C

Which gives us:

∫ x√(4-x²) dx
=
-1/3(4-x²)^3/2

Which, when we apply the limits 0 to 2, gives us 8/3.

Answer 2

4-x^2>=0 so -2<=x<=2
So you must calculate the integral between x= 0 and x=2 of
Intx*sqrt(4-x^2) dx
Put 4-x^2=z so -2xdx=dz
and we have the
-1/2Int sqrt(z) between z=4 and z=0
=-1/2*2/3( -4^3/2) = 1/3 *8 = 8/3 (B)