Question

Evaluate the following integral \[\int\limits_{0}^{2}{\sqrt{4-{{x}^{2}}}dx}\].

Answer 1

Hint: In this question, in order to evaluate the integral \[\int\limits_{0}^{2}{\sqrt{4-{{x}^{2}}}dx}\], we have to first substitute \[x=2\sin t\], then in the give integral the lower limit and upper limit of the variable \[x\] should be changed \[t\] by putting the value \[x=0\] and \[x=2\] in \[x=2\sin y\] to find the respective lower limit and upper limit of the variable when we are changing the variable from \[x\] to \[y\]. Also we have to determine the value of \[dx\] in terms of the variable \[y\] and \[dy\]. We will then evaluate the simplified integral in terms of variable \[y\].

Complete step by step answer:
Let \[I\] denote the integral \[\int\limits_{0}^{2}{\sqrt{4-{{x}^{2}}}dx}\].
That is, let \[I=\int\limits_{0}^{2}{\sqrt{4-{{x}^{2}}}dx}...........(1)\].
Now let us suppose that \[x=2\sin y.....(2)\].
Now on differentiate \[x=2\sin y\] where differentiation of \[\sin y\] is equals to \[\cos ydy\] in order to determine the value of \[dx\] in terms of the variable \[y\] and \[dy\], we will get
\[dx=2\cos ydy......(3)\].
Now we will evaluate the lower limit of the integral \[I\] by the value \[x=0\] in \[x=2\sin y\].
Putting the value the value \[x=0\] in \[x=2\sin y\] , we get
\[\begin{align}
  & 0=2\sin y \\
 & \Rightarrow \sin y=0 \\
\end{align}\]
Since we have \[\sin y=0\] when \[y=0\], thus the lower limit of the variable \[y\] is \[0\].
We will now calculate the upper limit of the integral \[I\] by the value \[x=2\] in \[x=2\sin y\].
Putting the value the value \[x=2\] in \[x=2\sin y\] , we get
\[\begin{align}
  & 1=2\sin y \\
 & \Rightarrow \sin y=1 \\
\end{align}\]
Since we have \[\sin y=1\] when \[y=\dfrac{\pi }{2}\], thus the lower limit of the variable \[y\] is \[\dfrac{\pi }{2}\].
Now on substituting equation (2) and equation (3) in equation (1) and by changing the lower limit and upper limit of variable \[y\] in equation (1), we will get
\[\begin{align}
  & I=\int\limits_{0}^{2}{\sqrt{4-{{x}^{2}}}dx} \\
 & =\int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{4-{{\left( 2\sin y \right)}^{2}}}2\cos ydy} \\
 & =\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \sqrt{4-4{{\sin }^{2}}y} \right)2\cos ydy} \\
 & =\int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{4\left( 1-{{\sin }^{2}}y \right)}2\cos ydy}
\end{align}\]
Using \[1-{{\sin }^{2}}y={{\cos }^{2}}y\] in the above integral, we get
\[\begin{align}
  & I=\int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{4\left( {{\cos }^{2}}y \right)}2\cos ydy} \\
 & =\int\limits_{0}^{\dfrac{\pi }{2}}{\left( 2\cos y \right)2\cos ydy} \\
 & =4\int\limits_{0}^{\dfrac{\pi }{2}}{{{\cos }^{2}}ydy}
\end{align}\]
 We will now use the identity \[2{{\cos }^{2}}\theta =1+\cos 2\theta \] in the above integral.

That is on substituting \[{{\cos }^{2}}y=\dfrac{1+\cos 2y}{2}\] in the above integral, we will have
\[\begin{align}
  & I=4\int\limits_{0}^{\dfrac{\pi }{2}}{{{\cos }^{2}}ydy} \\
 & =4\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{1+\cos 2y}{2} \right)dy} \\
 & =\dfrac{4}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\left( 1+\cos 2y \right)dy} \\
 & =2\int\limits_{0}^{\dfrac{\pi }{2}}{\left( 1+\cos 2y \right)dy}
\end{align}\]
Now we will evaluate the above integrals by parts,
\[\begin{align}
  & I=2\int\limits_{0}^{\dfrac{\pi }{2}}{\left( 1+\cos 2y \right)dy} \\
 & =2\int\limits_{0}^{\dfrac{\pi }{2}}{1dy}+2\int\limits_{0}^{\dfrac{\pi }{2}}{\cos 2ydy} \\
 & =2\left[ y \right]_{0}^{\dfrac{\pi }{2}}+2\left[ \dfrac{\sin 2y}{2} \right]_{0}^{\dfrac{\pi }{2}} \\
 & =2\left( \dfrac{\pi }{2}-0 \right)+\left( \sin 2\left( \dfrac{\pi }{2} \right)-\sin 0 \right) \\
\end{align}\]
Since with know that \[\sin \left( n\pi \right)=0\,\,\,\,\,\,\forall n\in Z\], thus we get
\[\begin{align}
  & I=2\left( \dfrac{\pi }{2}-0 \right)+\left( \sin 2\left( \dfrac{\pi }{2} \right)-\sin 0 \right) \\
 & =\pi +\left( 0-0 \right) \\
 & =\pi
\end{align}\]

Therefore we have \[\int\limits_{0}^{2}{\sqrt{4-{{x}^{2}}}dx}=\pi \].

Note: In this problem, while evaluating the integrals \[\int\limits_{0}^{2}{\sqrt{4-{{x}^{2}}}dx}\] we are changing the integral in terms of variable \[y\] using \[x=2\sin y\]. Please take care of the fact that while evaluating the integral in terms of variable \[y\] we have to change everything from variable \[x\] to \[y\] including the lower limit and the upper limit of the integral.