Find the area lying above the x - axis and under the parabola y = 4x – x2.
Given equations are:
x – axis ...... (1)
And y = 4x – x2 ...... (2)
⇒ y + 4 = – (x2 – 4x – 4) (adding 4 on both sides)
⇒ – (y + 4) = (x – 2)2
equation (2) represents a downward parabola with vertex at (2,4) and passing through (0,0) and (4,0) on the x – axis, A rough sketch is given as below: –
We have to find the area of the shaded region.
= shaded region OABO (as it is symmetrical about the x - axis)
(the area can be found by taking a small slice in each region of width Δx, then the area of that sliced part will be yΔx as it is a rectangle and then integrating it to get the area of the whole region)
(As x is between (0,4) and the value of y varies)
(as y = 4x – x2)
On integrating we get,
On applying the limits, we get,
Hence the area lying above the x - axis and under the parabola y = 4x – x2 is equal to square units.