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

A_{1} = (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

A_{2} = (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

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