Question

How do you evaluate the integral of\[\int{\left| \sin x \right|}dx\] from 0 to \[\dfrac{3\pi }{2}\].

Answer 1

Hint: In this problem, we have to evaluate the given integral within the given interval. We know that to solve these types of problems, we should know some integral formulas to apply in the problem and to get the final value. We know that the \[\sin x\] function is positive in the first and second quadrant and negative in the third and fourth quadrant, for this reason we can break the interval into two parts and can integrate the two parts with the separated limit values to get the answer.

Complete step by step answer:
We know that the given integral and interval is \[0\le x\le \dfrac{3\pi }{2}\]
\[\int{\left| \sin x \right|}dx\] ….. (1)
Now we can split the above integral and interval into two parts as the\[\sin x\] function is positive in the first and second quadrant and negative in the third and fourth quadrant.
\[\left| \sin x \right|=\sin x\] for \[0\le x\le \pi \]
\[\left| \sin x \right|=-\sin x\] for \[\pi \le x\le \dfrac{3\pi }{2}\].
Now we can rewrite the integral (1) into two parts, we get
\[\Rightarrow \int\limits_{0}^{\pi }{\sin xdx-\int\limits_{\pi }^{\dfrac{3\pi }{2}}{\sin xdx}}\]
Now we can integrate the above step and apply the limits, we get
\[\begin{align}
  & \Rightarrow \left[ -\cos x \right]_{0}^{\pi }-\left[ -\cos x \right]_{\pi }^{\dfrac{3\pi }{2}} \\
 & \Rightarrow \left[ -\cos x \right]_{0}^{\pi }+\left[ \cos x \right]_{\pi }^{\dfrac{3\pi }{2}} \\
\end{align}\]
\[\because \int{\sin xdx=-\cos x}\]
 We have to solve the above step using second fundamental theorem of calculus,
We know that,
\[\int\limits_{a}^{b}{f\left( x \right)}dx=\left[ F\left( x \right) \right]_{a}^{b}=F\left( b \right)-F\left( a \right)\]
We can apply this in the above step, we get
\[\Rightarrow \left[ -\cos \pi -\left( -\cos 0 \right) \right]+\left[ \cos \dfrac{3\pi }{2}-\cos \pi \right]\]
Now we can substitute the radian values for cosine, we get
\[\begin{align}
  & \Rightarrow \left[ -\left( -1 \right)-\left( -1 \right) \right]+\left[ 0-\left( -1 \right) \right] \\
 & \Rightarrow 2+1=3 \\
\end{align}\]
\[\because \cos 0=1,\text{ }\cos \pi =-1,\text{ }\cos \dfrac{3\pi }{2}=0\] .
Therefore, the value of \[\int{\left| \sin x \right|}dx\] from 0 to \[\dfrac{3\pi }{2}\] is 3.

Note:
Students should know that, to solve these types of problems, we have to know basic integration methods and formulas. We should also know some trigonometric values of sines and cosines to solve these types of problems. We should understand the concept of definite integral and formulas used to solve these types of problems.