Question

How do you find the area under one half period of \[y=2\sin 3x\]?

Answer 1

Hint: We have a sine wave function whose area under one half period is to be found. We will first find the time period of the given wave function using the formula \[\dfrac{2\pi }{b}\], where b is the coefficient of \[x\] in the given equation. Since the half time period is asked we will divide it by 2 and we get it as \[\dfrac{\pi }{3}\]. We have to find the area, so we will integrate the given equation under the interval \[0\to \dfrac{\pi }{3}\].

Complete step by step solution:
According to the given question, we have been given a sinusoidal wave function which is of the standard form \[y=a\sin (bx+c)\]. We have to find the area under the curve for a half time period.
We will first start by finding the time period,
We know that the time period of a wave is defined as the time taken for one complete cycle. So, we have the formula for time period as per the standard wave function:
\[\text{Time period}=\dfrac{2\pi }{b}\]
But we have to find the half time period, we have,
Half time period \[=\dfrac{\pi }{b}\], where b is the coefficient of \[x\].
We have, half time period \[=\dfrac{\pi }{3}\]
Now, in order to find the area under the curve we will have to integrate the given expression in the interval \[x=0\to \dfrac{\pi }{3}\].
Area \[=\int\limits_{0}^{\dfrac{\pi }{3}}{2\sin 3xdx}\]
We know that \[\int{\sin xdx=-\cos x}\], after integrating the function the angle \[3x\] will get differentiated as well.
 We get,
\[\Rightarrow Area=2\left[ \dfrac{-\cos 3x}{3} \right]_{0}^{\dfrac{\pi }{3}}\]
Applying the intervals in the function, we get,
\[\Rightarrow Area=\dfrac{-2}{3}\left[ \cos 3\left( \dfrac{\pi }{3} \right)-\cos 3(0) \right]\]
\[\Rightarrow Area=\dfrac{-2}{3}\left[ \cos \pi -\cos 0 \right]\]
\[\Rightarrow Area=\dfrac{-2}{3}\left[ -1-1 \right]\]
\[\Rightarrow Area=\dfrac{-2}{3}\left[ -2 \right]\]
\[\Rightarrow Area=\dfrac{4}{3}\] sq. units
Therefore, we get the area under the given expression for the one half time period is \[\dfrac{4}{3}\] sq. units.

Note: The formula for time period we have is for one complete cycle, so when it is asked for half the cycle or three-fourths, etc. necessary changes should be made. Also, the intervals should be applied on the function only after the function is integrated and not before. And the sequence for applying the limits are: upper limit – lower limit.