Question

Find the area bounded by the curve y = x(4 – x) and the x-axis from x = 0 to x = 5 as shown in the figure given above.

Answer 1

The curve is y = x(4 – x) = 4x – x²

It cuts the x-axis when y = 0:

x(4 – x) = 0

x = 0, 4

So, from x = 0 to x = 4, y ≥ 0 (area above x-axis) and from x = 4 to x = 5, y ≤ 0 (area below x-axis).

`A = int_0^4 y  dx + int_4^5 - y  dx`

= `int_0^4 x(4 - x)dx + int_4^5 x(x - 4)dx`

= `int_0^4 (4x - x^2)dx + int_4^5 (x^2 - 4x)dx`

= `[2x^2 - x^3/3]_0^4 + [x^3/3 - 2x^2]_4^5`

= `(2 xx 4^2 - 4^3/3) - (2 xx 0^2 - 0^3/3) + (5^3/3 - 2 xx 5^2) - (4^3/3 - 2 xx 4^2)`

= `(32 - 64/3) + (125/3 - 50) - (64/3 - 32)`

= `32/3 + (125 - 150)/3 - (64 - 96)/3`

= `32/3 - 25/3 + 32/3`

= `39/3`

= 13 sq. units