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Make a rough sketch of the graph of the function y = 4 – x2, 0 ≤ x ≤ 2 and determine the area enclosed by the curve, the x - axis and the lines x = 0 and x = 2.
Given equations are:
x – axis ...... (1)
x = 0 ...... (2)
x = 2 ...... (3)
And y = 4 – x2, 0 ≤ x ≤ 2 ...... (4)
⇒ y = – (x2 – 4) ⇒ x2 = – (y – 4)
equation (4) represents a downward parabola with vertex at (0,4) and passing through (2,0) and ( – 2,0) on x – axis, equation (3) represents a line parallel to y – axis at a distance of 2 units and equation (2) represents y - axis.
A rough sketch is given as below: -
We have to find the area of the shaded region.
= shaded region OABO
(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 (0,2) and the value of y varies)
(as y = 4 – x2 )
(as x0 = 1)
On integrating we get,
On applying the limits, we get,
Hence the area enclosed by the curve, the x - axis and the lines x = 0 and x = 2 is equal to square units.