Question

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

Answer 1

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

& 0=2\sin y \\\ & \Rightarrow \sin y=0 \\\ \end{aligned}$$ Since we have $$\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{aligned} & 1=2\sin y \\\ & \Rightarrow \sin y=1 \\\ \end{aligned}$$ Since we have $$\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{aligned} & 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{aligned}$$ Using $$1-{{\sin }^{2}}y={{\cos }^{2}}y$$ in the above integral, we get $$\begin{aligned} & 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{aligned}$$ We will now use the identity $$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{aligned} & 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{aligned}$$ Now we will evaluate the above integrals by parts, $$\begin{aligned} & 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{aligned}$$ Since with know that $$\sin \left( n\pi \right)=0\,\,\,\,\,\,\forall n\in Z$$, thus we get $$\begin{aligned} & 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{aligned}$$ **Therefore we have $$\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.