Find the area lying above the x - axis and under the parabola y = 4x – x^{2}.

Given equations are:

x – axis ...... (1)

And y = 4x – x^{2} ...... (2)

⇒ y + 4 = – (x^{2} – 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.

Required area

= 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 – x^{2})

On integrating we get,

On applying the limits, we get,

Hence the area lying above the x - axis and under the parabola y = 4x – x^{2} is equal to square units.

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