2.31: Stationary point and area under a semicircle curve

0606/21/O/N/20 — Question 11 · 11 marks

The equation of a curve is

y = x\sqrt{16 - x^{2}}

for

0 \le x \le 4

(a)

Find the exact coordinates of the stationary point of the curve.

[6]

(b)

Find

\dfrac{\mathrm{d}}{\mathrm{d}x}(16 - x^{2})^{\frac{3}{2}}

and hence evaluate the area enclosed by the curve

y = x\sqrt{16 - x^{2}}

and the lines

y = 0

x = 1

and

x = 3

[5]