Suppose that the region bounded by $y=x^4+1$, $x$-axis, $x=1$ and $x=2$, as shown in Figure 1, is revolved about the $y$-axis and we want to find the volume of the resulting solid.

Figure 1

You will find it difficult to apply the disk method or the washer method for this though it’s doable. Here we want to devise another method that can come in handy for this kind of problems. First let us take a look at Figure 2.

Figure 2

Imagine that each rectangle is rotated about the $y$-axis. Then we would obtain layers of cylindrical shells and the sum of the volumes of these cylindrical shells would approximate the volume of the solid in question. If we use infinitesimally thin rectangles, the sum of the volumes of cylindrical shells would be exactly the volume of the solid. The radius of the cylindrical shell at $x$ is $x$ since the axis of rotation is the $y$-axis ($x=0$), its height is $x^4+1$, and its infinitesimal thickness is $dx$. So the volume of the cylindrical shell at $x$ is $2\pi x(x^4+1)dx$. Therefore the volume of the solid is $$V=\int_1^2 2\pi x(x^4+1)dx=24\pi$$

Example. Find the volume of the solid formed by rotating the region enclosed by $$x=0,\ x=1,\ y=0,\ y=9+x^3$$ about the $y$-axis.

Solution.

Figure 3

The enclosed region is shown in Figure 3. Using the cylindrical shell method, the volume is \begin{align*}V&=\int_0^1 2\pi x(9+x^3)dx\\&=2\pi\int_0^1(9x+x^4)dx\\&=2\pi\left[\frac{9}{2}x^2+\frac{x^5}{5}\right]_0^1\\&=\frac{47}{5}\pi\end{align*}

Remark. The volume can be also found by the disk/washer method but its solution is more complicated than the one by cylindrical shell method. In order to use the disk/washer method, we need to break the enclosed region into two parts as shown in Figure 4.

Figure 4

When rotated about the $y$-axis, the region in red will give rise to a solid cylinder with the radius 1 and the height 9. Its volume is $9\pi$. For the solid formed by rotating the region in blue, using the washer method, its volume is given by \begin{align*}\int_9^{10}\pi [1^2-((y-9)^{\frac{1}{3}})^2]dy&=\pi\int_9^{10}[1-(y-9)^{\frac{2}{3}}]dy\\&=\pi\left[y-\frac{3}{5}(y-9)^{\frac{5}{3}}\right]_9^{10}\\&=\frac{2}{5}\pi\end{align*} Therefore, the volume $V$ of the solid in question is given by $$V=9\pi+\frac{2}{5}\pi=\frac{47}{5}\pi$$

More generally we have

Cylindrical Shell Method

Suppose that the region under the curve $y=f(x)$, $a\leq x\leq b$ is revolved about the line $x=L$. Then the volume $V$ of the resulting solid is obtained by $$V=\left\{\begin{array}{ccc}\int_a^b 2\pi (x-L)f(x)dx & \mbox{if} & L\leq a\\\int_a^b 2\pi (L-x)f(x)dx & \mbox{if} & L\geq b\end{array}\right.$$

Example. The region enclosed by the $x$-axis and the parabola $y=3x-x^2$ is revolved about the vertical line $x=-1$. Find the volume of the resulting solid.

Solution. The region (in red) and the axis of rotation (in blue) are shown in Figure 5.

Figure 5

By cylindrical shell method, the volume $V$ is given by $$V=\int_0^3 2\pi(x+1)(3x-x^2)dx=\frac{45\pi}{2}$$

Example. The region bounded by the curve $y=\sqrt{x}$, the $x$-axis and the line $x=4$ is revolved about the $y$-axis. Find the volume of the resulting solid.

Solution. Figure 6 shows the region.

Figure 6

Since the axis of rotation is $x=0$, by cylindrical shell method the volume $V$ is given by $$V=\int_0^42\pi x\sqrt{x}dx=\frac{128\pi}{5}$$

Note that while cylindrical shell method is simpler, it is also easy to find the volume using washer method. $$V=\int_0^2\pi[4^2-(y^2)^2]dy=\int_0^2(16-y^4)dy=\frac{128\pi}{5}$$

Example. The same region in the previous example is revolved about the $x$-axis this time. Find the volume of the resulting solid using cylindrical shell method.

Solution. The solid and a cylindrical shell are shown in Figure 7.

Figure 7

The shell radius is $y$ and the shell height $4-y^2$. Hence the volume is $$V=\int_0^2 2\pi y(4-y^2)dy=8\pi$$

Note that it would actually be easier to use disk method for this problem. Using disk method the volume $V$ is given by $$V=\int_0^4\pi(\sqrt{x})^2dx=\int_0^4\pi xdx=8\pi$$

Let $y=f(x)$ and $y=g(x)$ be two continuous functions on a closed interval $[a,b]$ such that $f(x)\geq g(x)$ on $[a,b]$ as seen in Figure 1.

Figure 1

The area $A$ of the region between the curves $f(x)$ and $g(x)$ is \begin{align*}A&=\int_a^b f(x)dx-\int_a^b g(x)dx\\&=\int_a^b[f(x)-g(x)]dx\end{align*}

Example. Find the area of the region enclosed by the parabola $y=2-x^2$ and the line $y=-x$.

Solution. First we need to find the $x$-coordinates of the points at which the two curves meet. Set $2-x^2=-x$. Then the quadratic equation $x^2-x-2=0$ has two solutions $x=-1,2$. From Figure 2

Figure 2

we see that the area is \begin{align*}A&=\int_{-1}^2[(2-x^2)-(-x)]dx\\&=\int_{-1}^2(2+x-x^2)dx\\&=\frac{9}{2}\end{align*}

Example. Find the area of the region in the first quadrant that is bounded above by $y=\sqrt{x}$ and below by the $x$-axis and the line $y=x-2$.

Solution. To find the $x$-coordinates of the points at which $y=\sqrt{x}$ and $y=x-2$ meet, set $\sqrt{x}=x-2$. Squaring this equation, we obtain the quadratic equation $x^2-5x+4=0$ whose solutions are $x=-1$ and $x=4$. Since we are considering only the first quadrant $x=-1$ is not our concern. The bounded region is shown in Figure 3.

Figure 3

Note that on $[0,2]$ the part of the region is bounded by $y=\sqrt{x}$ and the $x$-axis and on $[2,4]$ the part of the reagion is bounded by $y=\sqrt{x}$ and $y=x-2$. Hence, the area is \begin{align*}A&=\int_0^2\sqrt{x}dx+\int_2^4[\sqrt{x}-(x-2)]dx\\&=\frac{10}{3}\end{align*}

If $x=f(y)$ and $x=g(y)$ are two continuous functions on a closed interval $[c,d]$ such that $f(y)\geq g(y)$ as seen in Figure 4,

Figure 4

then the area $A$ bounded by $f(y)$ and $g(y)$ is given by $$A=\int_c^d[f(y)-g(y)]dy$$

Example. Redo the previous example by integrating with respect to $y$.

Solution. The same region in Figure 3 can be considered as the region bounded by $x=y^2$ ($y\geq 0$ and $x=y+2$. Hence, $$A=\int_0^2[y+2-y^2]dy=\frac{10}{3}$$

Often the fastest way to find an area may be to combine calculus with geometry. The same region in Figure 3 can be viewed as in Figure 5.

Figure 5

The area of the right triangle in blue is $\frac{1}{2}\cdot 2\cdot 2=2$ hence, the area of the region is $$A=\int_0^4\sqrt{x}dx-2=\frac{10}{3}$$

Disk Method

Consider the solid obtained by rotating a region bounded by the function $y=f(x)$ and the $x$-axis, $a\leq x\leq b$.

Figure 1

The cross section of the resulting sold at $x$ is a disk of radius $f(x)$ as seen in Figure 1 and so its area is $A(x)=\pi [f(x)]^2$. Hence the volume of the solid of revolution is \begin{equation}\label{eq:diskmethod}V=\pi\int_a^b [f(x)]^2dx\end{equation}

Example. The region between the curve $y=\sqrt{x}$, $0\leq x\leq 4$ and the $x$-axis is revolved about the $x$-axis. Find the volume of the resulting solid.

Solution. Figure 2 shows the region

Figure 2

and Figure 3 shows the resulting solid.

Figure 3

The volume $V$ is given by $$V=\pi\int_0^4[\sqrt{x}]^2dx=\pi\int_0^4 xdx=8\pi$$

Example. Find the volume of the solid generated by revolving the region bounded by $y=\sqrt{x}$ and the lines $y=1$, $x=4$ about the line $y=1$.

Solution. Figure 4 shows the region

Figure 4

and Figure 5 shows the resulting solid.

Figure 5

Note that the radius of the cross section at $x$ is $\sqrt{x}-1$ and so the volume is given by $$V=\pi\int_1^4[\sqrt{x}-1]^2dx=\frac{7}{6}\pi$$

Example. Find the volume of the solid obtained by revolving the region between the $y$-axis and the curve $x=\frac{2}{y}$, $1\leq y\leq 4$ about the $y$-axis.

Solution. Figure 6 shows the region

Figure 6

and Figure 7 shows the resulting solid.

Figure 7

The area of the cross section at $y$ is $A(y)=\pi\left(\frac{2}{y}\right)^2$. Hence the volume is $$V=\int_1^4 A(y)dy=\pi\int_1^4\left(\frac{2}{y}\right)^2dy=3\pi$$

Consider a solid each of which parallel cross sections has its area represented by the function $A(x)$, $a\leq x\leq b$. The volume element is then given by $dV=A(x)dx$. Hence the volume of the solid is obtained by the integral \begin{equation}\label{eq:volslice}V=\int_{x=a}^{x=b}dV=\int_a^b A(x)dx\end{equation}

Example. Find the volume of a pyramid whose base is a square with side $L$ and whose height is $h$.

Solution. See Figure 1.

Figure 1

We find the ratio $x$ to $h$ $\frac{x}{h}=\frac{\frac{s}{2}}{{\frac{L}{2}}}=\frac{s}{L}$. So we have $s=\frac{L}{h}x$. (Another observation is the line segment $\overline{OP}$ has slope $\frac{\frac{L}{2}}{h}=\frac{L}{2h}$ so the equation of line through $\overline{OP}$ is $y=\frac{L}{2h}x$ and for $y=\frac{s}{2}$ we get $s=\frac{L}{h}x$.) The area $A(x)$ of the cross section at $x$ is $A(x)=s^2=\frac{L^2}{h^2}x^2$. Hence $$V=\int_0^h A(x)dx=\frac{L^2}{h^2}\int_0^h x^2dx=\frac{1}{3}L^2h$$

Example. Show that the volume of a right circular cone whose circular base has radius $r$ and whose height is $h$ is $V=\frac{1}{3}\pi r^2h$.

Solution. Left as an exercise.

Example. Show that the volume of a sphere of radius $r$ is $V=\frac{4}{3}\pi r^3$.

Solution. See Figure 2.

Figure 2

$y=\sqrt{r^2-x^2}$ so $A(x)=\pi y^2=\pi (r^2-x^2)$. Hence the volume $V$ is \begin{align*}V&=\int_{-r}^r A(x)dx\\&=\int_{-r}^r \pi (r^2-x^2)dx\\&=2\pi\int_0^r (r^2-x^2)dx\\&=2\pi\left[r^2x-\frac{x^3}{3}\right]_0^r\\&=\frac{4}{3}\end{align*}

Example. Consider a solid with a circular base of radius 1 whose parallel cross secxtions perpendicular to the base are equilateral triangles. Find the volume of the solid.

Solution. See Figure 3.

Figure 3

From the picture we see that $$A(x)=\frac{1}{2}(2y)(\sqrt{3}y)=\sqrt{3}y^2=\sqrt{3}(1-x^2)$$ Hence the volume $V$ is given by $$V=\int_{-1}^1 A(x)dx=\sqrt{3}\int_{-1}^1(1-x^2)dx=2\sqrt{3}\int_0^1(1-x^2)dx=\frac{4\sqrt{3}}{3}$$

Example. A Curved wedge is cut from a cylinder of radius 3 by two planes. One plane is perpendicular to the axis of the cylinder. The other plane is crossing the first plane at a $45^\circ$ angle at the center of the cylinder.

Solution. See Figure 4.

Figure 4

The cross section at $x$ is the rectangle in blue. It’s area is $$A(x)=\mbox{length}\times\mbox{width}=2\sqrt{9-x^2}\times x=2x\sqrt{9-x^2}$$ Hence the volume $V$ is given by \begin{align*}V&=\int_0^3 2x\sqrt{9-x^2}dx\\&=-\int_9^0\sqrt{u}du\ (u=9-x^2)\\&=\int_0^9\sqrt{u}du\\&=\frac{2}{3}[u^{\frac{3}{2}}]_0^9\\&=18\end{align*}

Linear Approximation of $f(x)$ at $a$

As seen in differential calculus, when $x$ is near $a$ the function $f(x)$ can be approximated by the tangent line to $y=f(x)$ at $a$.  (For details see here.) \begin{equation}\label{eq:linapprox}T_1(x)=f(a)+f'(a)(x-a)\end{equation} Note that $T_1(a)=f(a)$ and $T_1′(a)=f'(a)$.

Quadratic Approximation of $f(x)$ at $a$

If the graph of $y=f(x)$ is curved a lot near $a$, a polynomial of higher degree may be preferred for a better approximation, for instance a quadratic polynomial. Let us call such polynomial $T_2(x)$. We require that $T_2(x)$ satisfies $T_2(a)=f(a)$, $T_2′(a)=f'(a)$ and $T_2^{\prime\prime}(a)=f^{\prime\prime}(a)$. We can find $T_2(x)$ be setting $$T_2(x)=f(a)+f'(a)(x-a)+cf^{\prime\prime}(a)(x-a)^2$$ Then the condition $T_2^{\prime\prime}(a)=f^{\prime\prime}(a)$ implies that $c=\frac{1}{2}$. Hence \begin{equation}\label{eq:quadapprox}T_2(x)=f(a)+f'(a)(x-a)+\frac{f^{\prime\prime}(a)}{2!}(x-a)^2\end{equation} There is a reason for the appearance of the factorial notation in the leading coefficient.

Example. Approximation for $\ln x$.

1. Find the linear approximation to $f(x)=\ln x$ at $a=1$.
2. Find the quadratic approximation to $f(x)=\ln x$ at $a=1$.
3. Use these approximations to estimate the value of $\ln(1.05)$.

Solution. $f'(x)=\frac{1}{x}$ and $f^{\prime\prime}(x)=-\frac{1}{x^2}$. So

1. $T_1(x)=x-1$ and
2. $T_2(x)=(x-1)-\frac{1}{2}(x-1)^2$.
3. $T_1(1.05)=0.05$ and $T_2(1.05)=0.04875$. For comparison, the actual value of $\ln(1.05)$ is $0.04879016417\cdots$.

The graphs of y=ln(x) (in black), T_1(x)=x-1 (in blue) and T_2(x)=(x-1)-(x-2)^2/2 (in red)

The $n$-th Order Approximation of $f(x)$ at $a$

The $n$-th order approximation is given by the Taylor polynomial $T_n(x)$ centered at $a$ \begin{equation}\label{eq:taylorpolynomial}T_n(x)=f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f^{\prime\prime}(a)}{2!}(x-a)^2+\cdots+\frac{f^{(n)}(a)}{n!}(x-a)^n\end{equation}

Example. Find the Taylor polynomials $T_1,T_2,\cdots,T_7$ for $f(x)=\sin x$ at $a=0$.

Solution. $T_1(x)=T_2(x)=x$, $T_3(x)=T_4(x)=x-\frac{x^3}{3!}$, $T_5(x)=T_6(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}$, $T_7(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}$.

Example. Use Taylor polynomials of order $n=0,1,2,3$ to approximate $\sqrt{18}$.

Solution. There are two things we need to choose $f(x)$ and $a$. Obviously the function we need to use is $f(x)=\sqrt{x}$. How do we choose $a$ then? There are two things to consider for choosing a suitable $a$. One is $a$ has to be close to 18 and the other is $f(a)=\sqrt{a}$ is a number that can be calculated without a calculator. If we are allowed to use a calculator, the point of doing an approximation is moot. Besides, this is how people were able to calculate a number like $\sqrt{18}$ along time ago when calculators did not exist. In fact, calculators or computers cannot calculate the exact $\sqrt{18}$. What they can do is also approximations using the Taylor polynomials. With the two things in mind, the suitable choice for $a$ would be $a=16$. $f'(x)=\frac{1}{2\sqrt{x}}$, $f^{\prime\prime}(x)=-\frac{1}{4x\sqrt{x}}$, $f^{\prime\prime\prime}(x)=\frac{3}{8x^2\sqrt{x}}$. So, \begin{align*}T_0(x)&=\sqrt{16}=4\\T_1(x)&=4+\frac{1}{8}(x-16)\\T_2(x)&=4+\frac{1}{8}(x-16)-\frac{1}{512}(x-16)^2\\T_3(x)&=4+\frac{1}{8}(x-16)-\frac{1}{512}(x-16)^2+\frac{1}{16,384}(x-16)^3\end{align*} Hence, $T_0(18)=4$, $T_1(18)=4.25$, $T_2(18)=4.242188$, $T_3(18)=4.242676$. The actual value is $\sqrt{18}=4.242640686\cdots$.

When a function $f(x)$ is approximated by a Taylor polynomial $T_n(x)$, the error bound must also be taken into consideration because it can tell us about the accuracy of the approximation. The error is given by the remainder \begin{equation}\label{eq:remainder}R_n(x)=f(x)-T_n(x)\end{equation}

Theorem (Taylor’s Theorem). Let $f$ have continuous derivatives up to $f^{(n+1)}$ on an open interval $I$ containing $a$. Then for all $x$ in $I$, $$f(x)=T_n(x)+R_n(x),$$ where $T_n(x)$ is the $n$-th order Taylor polynomial for $f$ centered at $a$ and the remainder $R_n$ is \begin{equation}\label{eq:remainder2}R_n(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1}\end{equation} for some $\xi$ between $x$ and $a$.

Remark. Recall the Mean Value Theorem: If $f(x)$ is continuous on $[a,x]$ and is differentiable on $(a,x)$, then there exists $a<\xi<x$ such that $$\frac{f(x)-f(a)}{x-a}=f'(\xi)$$ or $$f(x)=f(a)+f'(\xi)(x-a)$$ Hence we see that the Taylor’s theorem is a generalization of the Mean Value Theorem.

\eqref{eq:remainder2} can be used to obtain the maximum error bound \begin{equation}\label{eq:errorbd}|R_n(x)|=|f(x)-T_n(x)|\leq M\frac{|x-a|^{n+1}}{(n+1)!}\end{equation}where $|f^{(n+1)}(\xi)|\leq M$ for all $\xi$ between $a$ and $x$.

Example.

1. What is the maximum error possible in using the approximation $$\sin x\approx x-\frac{x^3}{3!}+\frac{x^5}{5!}$$ when $-0.3\leq x\leq 0.3$? Use this approximation to find $\sin 12^\circ$, correct to six decimal places.
2. For what values of $x$ is this approximation accurate to within 0.00005?

Solution.

1. Recall from the second example above $x-\frac{x^3}{3!}+\frac{x^5}{5!}=T_5(x)=T_6(x)$ but we regard it as $T_6(x)$. The reason is that the error $R_6$ is smaller than $R_5$. (This choice is also consistent with the alternating series estimate because $-\frac{x^7}{7!}$ comes after the term $\frac{x^5}{5!}$. See below.) Since $|f^{(7)}(\xi)|=|\cos(\xi)|\leq 1$ for all $\xi$ between 0 and $x$, from \eqref{eq:errorbd}, we have $|R_6(x)|\leq\frac{|x|^7}{7!}=\frac{|x|^7}{5040}$. Since $|x|\leq 0.3$, $|R_6|\leq\frac{(0.3)^7}{5040}\approx 4.3\times 10^{-8}$. Thus the maximum possible error is $4.3\times 10^{-8}$. Note that $T_6(x)$ is an alternating series, so it is actually easier to use the remainder estimate for alternating series $$|R_6|\leq a_7=\frac{|x|^7}{7!}=\frac{|x|^7}{5040}$$ \begin{align*}\sin 12^\circ&=\sin\left(\frac{12\pi}{180}\right)\\&=\sin\left(\frac{\pi}{15}\right)\\&\approx\frac{\pi}{15}-\frac{\left(\frac{\pi}{15}\right)^3}{3!}+\frac{\left(\frac{\pi}{15}\right)^5}{5!}\\&\approx 0.207911694\end{align*} Using the maximum possible error, $$\sin 12^\circ=T_6\left(\frac{\pi}{15}\right)+R_6=0.207911694+0.0000000043=0.207911737$$ So $\sin 12^\circ$ correct to six decimal places is 0.207911.
2. $\frac{|x|^7}{5040}<0.00005$ so $|x|^7<0.00005\times 5045=0.252$. Hence, $|x|<(0.252)^{\frac{1}{7}}\approx 0.821$.

Example.

1. Approximate $f(x)=\root 3\of{x}$ by a Taylor polynomial of order (degree) 2 at $a=8$.
2. How accurate is this approximation when $7\leq x\leq 9$?

Solution.

1. First we find the following derivatives. \begin{align*}f'(x)&=\frac{1}{3}x^{-\frac{2}{3}}\\f^{\prime\prime}(x)&=-\frac{2}{9}x^{-\frac{5}{3}}\\f^{\prime\prime\prime}(x)&=\frac{10}{27}x^{-\frac{8}{3}}\end{align*} Now $$\root 3\of{x}\approx T_2(x)=2+\frac{1}{12}(x-8)-\frac{1}{288}(x-8)^2$$
2. Note that the Taylor polynomial is not alternating when $x<8$. So we use \eqref{eq:remainder2} instead for the remainder estimation.\begin{align*}R_2(x)&=\frac{f^{\prime\prime\prime}(\xi)}{3!}(x-8)^3\\&=\frac{10}{27}\xi^{-\frac{8}{3}}\frac{(x-8)^3}{3!}\\&=\frac{5}{81}\frac{(x-8)^3}{\xi^{\frac{8}{3}}}\end{align*} If $7\leq x\leq 9$,  $-1\leq x-8\leq 1$. Since $\xi>7$, $\xi^{\frac{8}{3}}>7^{\frac{8}{3}}>179$. Hence, $$|R_2(x)|\leq\frac{5}{81}\frac{|x-8|^3}{\xi^{\frac{8}{3}}}<\frac{5}{81}\cdot\frac{1}{179}\approx 0.00034485<0.0004$$

A Physical Application

In Einstein’s theory of special relativity, the mass of an object moving with velocity $v$ is \begin{equation}\label{eq:relmass}m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}\end{equation} where $m_0$ is the mass of the object at rest and $c$ is the speed of light in vacuum spacetime. $m$ in \eqref{eq:relmass} is called the relativistic mass. The kinetic energy is the difference between its total energy and its energy at rest \begin{equation}\label{eq:kenergy}K=mc^2-m_0c^2\end{equation}

1. Show that when $v\ll c$ (this means $v$ is very small compared with $c$), $K=\frac{1}{2}m_0v^2$.
2. Use Taylor’s formula to estimate the difference in these expressions for $K$ when $|v|\leq 100$ m/s.

Solution.

1. First note that \begin{align*}(1+x)^{-\frac{1}{2}}&=1-\frac{1}{2}x+\frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2!}x^2+\frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{3!}x^3+\cdots\\&=1-\frac{1}{2}x+\frac{3}{8}x^2-\frac{5}{16}x^3+\cdots\end{align*} (For details on how to obtain this read my note on Taylor series here, especially under The Binomial Series.) Thus, \begin{align*}K&=\frac{m_0c^2}{\sqrt{1-\frac{v^2}{c^2}}}-m_0c^2\\&=m_0c^2\left[\left(1-\frac{v^2}{c^2}\right)^{-\frac{1}{2}}-1\right]\\&=m_0c^2\left[\left(1+\frac{1}{2}\frac{v^2}{c^2}+\frac{3}{8}\frac{v^4}{c^4}+\frac{5}{16}\frac{v^6}{c^6}+\cdots\right)-1\right]\\&=m_0c^2\left(\frac{1}{2}\frac{v^2}{c^2}+\frac{3}{8}\frac{v^4}{c^4}+\frac{5}{16}\frac{v^6}{c^6}+\cdots\right)\end{align*} Hence, if $v\ll c$ then $K\approx\frac{1}{2}m_0v^2$.
2. $R_1(x)=\frac{f^{\prime\prime}(\xi)}{2!}x^2$ where $f(x)=m_0c^2[(1+x)^{-\frac{1}{2}}-1]$ with $x=-\frac{v^2}{c^2}$. $f^{\prime\prime}(x)=\frac{3}{4}m_0c^2(1+x)^{-\frac{5}{2}}$ so $R_1(x)=\frac{3m_0c^2}{8(1+\xi)^{\frac{5}{2}}}\cdot\frac{v^4}{c^4}$ where $-\frac{v^2}{c^2}<\xi<0$. With $c=3\times 10^8$ m/s and $|v|\leq 100$ m/s. we obtain \begin{align*}R_1(x)&\leq\frac{3m_0(9\times 10^{16})}{8\left(1-\frac{100^2}{c^2}\right)^{\frac{5}{2}}}\left(\frac{100}{c}\right)^4\\&<(4.17\times 10^{-10})m_0\end{align*}