Question

The equation y = 4 – x² represents a parabola.

1. Make a rough sketch of the graph of the given function.   [1]
2. Determine the area enclosed between the curve, the x-axis, the lines x = 0 and x = 2.   [2]
3. Hence, find the area bounded by the parabola and the x-axis.   [1]

Answer 1

a. Parabola y = 4 – x²

At x = 0, y = 4

The vertex of parabola is at (0, 4).

To find x-intercepts

Put y = 0, wet

0 = 4 – x²

⇒ x² = 4

⇒ x = ± 2

So, the x-intercepts are (–2, 0) and (2, 0).

b. Area between the curve, x-axis, x = 0 and x = 2.

Area = `int_0^2 |y| dx`

= `int_0^2 |4 - x^2| dx`

= `[4x - x^3/3]_0^2`

= `[4(2) - (2)^3/3] - (0)`

= `8/1 - 8/3`

= `(24 - 8)/3`

= `16/3` sq. units

c. Area bounded by the parabola and the x-axis = 2 × Area between the curve, x-axis, x = 0 and x = 2

Area = `2 xx 16/3`

= `32/2` sq. units