Question

How do you calculate the arc length of the curve $y={{x}^{2}}$ from $x=0$ to $x=4$?

Answer 1

Arc length is basically the distance between two points but along a particular curve. Thus, we shall apply the formula of calculating the arc lengths of curve which is given as $L=\int\limits_{a}^{b}{\sqrt{1+{{\left( {f}'\left( x \right) \right)}^{2}}}.dx}$. In accordance with it, we shall first find the derivative of the given function and then perform definite integration to apply the limits.

Complete step by step solution:
Given the curve $y={{x}^{2}}$ whose arc length is to be found from $x=0$ to $x=4$.
The arc length of a curve is given as
$L=\int\limits_{a}^{b}{\sqrt{1+{{\left( {f}'\left( x \right) \right)}^{2}}}.dx}$
Where,
${f}'\left( x \right)=$ First integral of the given curve whose arc length is to be calculated
$b=$ Rightmost point of the defined arc which becomes the upper limit of our definite integral
$a=$ Leftmost point of the defined arc which becomes the lower limit of our definite integral
We shall first differentiate the given curve $y={{x}^{2}}$ or $f\left( x \right)={{x}^{2}}$ with respect to x.
${f}'\left( x \right)=\dfrac{d}{dx}f\left( x \right)$
$\Rightarrow {f}'\left( x \right)=\dfrac{d}{dx}\left( {{x}^{2}} \right)$
Applying the basic property of derivatives, $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$, we get
$\Rightarrow {f}'\left( x \right)=2\left( {{x}^{2-1}} \right)$
$\Rightarrow {f}'\left( x \right)=2x$
We see that the rightmost point of our defined arc is $x=4$, thus the upper limit is equal to 4 and the leftmost point of our defined arc is $x=0$, thus the lower limit is equal to 0.
$\Rightarrow L=\int\limits_{0}^{4}{\sqrt{1+{{\left( 2x \right)}^{2}}}.dx}$
$\Rightarrow L=\int\limits_{0}^{4}{\sqrt{1+4{{x}^{2}}}.dx}$
Taking 4 common and then out of the square root as 2, we get
$\Rightarrow L=\int\limits_{0}^{4}{\sqrt{4\left( \dfrac{1}{4}+{{x}^{2}} \right)}.dx}$
$\Rightarrow L=\int\limits_{0}^{4}{2\sqrt{\left( \dfrac{1}{4}+{{x}^{2}} \right)}.dx}$
$\Rightarrow L=2\int\limits_{0}^{4}{\sqrt{\left( {{\left( \dfrac{1}{2} \right)}^{2}}+{{x}^{2}} \right)}.dx}$
In order to integrate this function of the form of a special integral, we shall make use of the formula
\Rightarrow L=\left. 2\left\\{ \dfrac{1}{2}\left[ x\sqrt{{{\left( \dfrac{1}{2} \right)}^{2}}+{{x}^{2}}}+{{\left( \dfrac{1}{2} \right)}^{2}}\log \left| x+\sqrt{{{\left( \dfrac{1}{2} \right)}^{2}}+{{x}^{2}}} \right| \right] \right\\} \right|_{1}^{4}
$\Rightarrow L=\left. \left[ x\sqrt{\dfrac{1}{4}+{{x}^{2}}}+\dfrac{1}{4}\log \left| x+\sqrt{\dfrac{1}{4}+{{x}^{2}}} \right| \right] \right|_{1}^{4}$
$\Rightarrow L=\left. \left[ x\sqrt{\dfrac{1}{4}+{{x}^{2}}}+\dfrac{1}{4}\log \left| x+\sqrt{\dfrac{1}{4}+{{x}^{2}}} \right| \right] \right|_{1}^{4}$
Applying the limits of integration, we get
$\Rightarrow L=\left[ 4\sqrt{\dfrac{1}{4}+{{4}^{2}}}+\dfrac{1}{4}\log \left| 4+\sqrt{\dfrac{1}{4}+{{4}^{2}}} \right| \right]-\left[ 1\sqrt{\dfrac{1}{4}+{{1}^{2}}}+\dfrac{1}{4}\log \left| 1+\sqrt{\dfrac{1}{4}+{{1}^{2}}} \right| \right]$
We know that ${{4}^{2}}=16$ and ${{1}^{2}}=1$.
$\Rightarrow L=\left[ 4\sqrt{\dfrac{1}{4}+16}+\dfrac{1}{4}\log \left| 4+\sqrt{\dfrac{1}{4}+16} \right| \right]-\left[ 1\sqrt{\dfrac{1}{4}+1}+\dfrac{1}{4}\log \left| 1+\sqrt{\dfrac{1}{4}+1} \right| \right]$
$\Rightarrow L=\left[ 4\sqrt{\dfrac{65}{4}}+\dfrac{1}{4}\log \left| 4+\sqrt{\dfrac{65}{4}} \right| \right]-\left[ 4\sqrt{\dfrac{5}{4}}+\dfrac{1}{4}\log \left| 1+\sqrt{\dfrac{5}{4}} \right| \right]$
$\Rightarrow L=\left( 4 \sqrt{\dfrac{65}{4}}-\sqrt{\dfrac{5}{4}} \right)+\dfrac{1}{4}\left( \log \left| 4+\sqrt{\dfrac{65}{4}} \right|-\log \left| 1+\sqrt{\dfrac{5}{4}} \right| \right)$
$\Rightarrow L=\left( 2\sqrt{65}-{\dfrac{\sqrt{5}}{2}} \right)+\dfrac{1}{4}\left( \log \left| 4+\dfrac{\sqrt{65}}{2} \right|-\log \left| 1+\dfrac{\sqrt{5}}{2} \right| \right)$
Applying the property of logarithms, $\log a-\log b=\log \dfrac{a}{b}$.
$\Rightarrow L=\left( 2\sqrt{65}-{\dfrac{\sqrt{5}}{2}} \right)+\dfrac{1}{4}\log \left| \dfrac{8+\sqrt{65}}{2+\sqrt{5}} \right|$
Therefore, the arc length of the curve $y={{x}^{2}}$ from $x=0$ to $x=4$ is $\Rightarrow L=\left( 2\sqrt{65}-{\dfrac{\sqrt{5}}{2}} \right)+\dfrac{1}{4}\log \left| \dfrac{8+\sqrt{65}}{2+\sqrt{5}} \right|$ units.

Note: We usually use definite integrals to find the area defined by some curves along with some other boundaries. However, we use definite integration for finding the arc length or area under a particular curve because the fundamental concept of integrals is breaking curves into infinitely small parts that we can approximate with lines and rectangles and then we could take the infinite sum of those infinitely small parts.