Show that the areas under the curves y = sin x and y = sin 2x between x = 0 and x = π/3 are in the ration 2:3.

Question

Show that the areas under the curves y = sin x and y = sin 2x between x = 0 and x = \(\frac{\pi}{3}\) are in the ration 2:3.

Answer 1

Given equations are:

y = sin x …..(i)

y = sin 2x …..(ii)

x = 0 ……(iii)

x = \(\frac{\pi}{3}\) .....(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 x = \(\frac{\pi}{3}\) 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)

On integrating we get,

On applying the limits we get

The area under the curve y = sin 2x , x = 0 and x = \(\frac{\pi}{3}\) 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)

On integrating we get,

So the ratio of the areas under the curves y = sin x and y = sin 2x between x = 0 and x = \(\frac{\pi}{3}\) are

Hence showed