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Show that the areas under the curves y = sin x and y = sin 2x between x = 0 and are in the ration 2:3.
Given equations are:
y = sin x …..(i)
y = sin 2x …..(ii)
x = 0 ……(iii)
x = …..(iv)
A table for values of y = sin x and y = sin 2x is: -
A rough sketch of the curves is given below: -
The area under the curve y = sin x , x = 0 and is
A1 = (area of the region OPBCA)
(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
and the value of y varies)
(as y = sin x)
On integrating we get,
On applying the limits we get
The area under the curve y = sin 2x , x = 0 and is
A2 = (area of the region OABCO)
(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
and the value of y varies)
(as y = sin 2x)
On integrating we get,
On applying the limits we get
So the ratio of the areas under the curves y = sin x and y = sin 2x between x = 0 and are
Hence showed