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: -


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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


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