Q.19 - Calculus

Question

(a) Explain why the definite integral $I={\int }_0^{0.5}\sin \left(x^2\right)dx$ exists. 

(b) Explain how to use Simpson's method to approximate $I$ to within $0.001$ of its actual value.

(c) Use substitution in the Maclaurin series for $\sin x$ to find a Maclaurin series for $\sin \left(x^2\right)$.

(d) Explain how to use Theorems $7.38$ and $8.12$
 to approximate $I$to within $0.001$ of its actual value.

Step-by-Step Solution

Answer

(a)The definite  integral $I$ exists because the function $\sin \left(x^2\right)$ is continuous in the interval $\left[0,0.5\right]$

(b) After using the Simpson's methods, the actual value  will be $0.00057$

(c)The Maclaurin series of $Sin\left(x^2\right)$ is $\sum _{k=0}^{\infty }\frac{(-1{)}^k}{(2k+1)!}{x}^{4k+2}$

(d)The actual value of the integral is $0.0416$ after using two theorems.

Part (a) Step 1: Given information

The integral is $I={\int }_0^{0.5}\sin \left(x^2\right)dx$

Part (a) Step 2: Calculation

The function of the integral  $\sin \left(x^2\right)$ is continuous in the interval  $\left[0,0.5\right]$.So, the definite integral exists.

Part (b) Step 1: Given information

The integral is $I={\int }_0^{0.5}\sin \left(x^2\right)dx$

Part (b) Step 2: Calculation

To use the Simpson's approximation to approximate the integral I within $0.001$ of its actual value, let us use the formula

${\int }_a^{b}f\left(x\right)dx\approx \frac{\Delta x}{3}\left[f\left(x_0\right)+4f\left(x_1\right)+2f\left(x_2\right)+\dots +2f\left(x_{n-2}\right)+4f\left(x_{n-1}\right)+f\left(x_n\right)\right]$

Where $\Delta x=\frac{b-a}{n}$

In the integral width="219" style="max-width: none; vertical-align: -19px;" $I={\int }_0^{0.5}\sin \left(x^2\right)dx,a=0,b=0.5$and width="37" style="max-width: none; vertical-align: -4px;" $n=3$ (since we need to approximate the integral / to within 0.001 of its actual value)

Therefore,

width="274" style="max-width: none; vertical-align: -15px;" $I\approx \frac{\Delta x}{3}\left[f\left(x_0\right)+4f\left(x_1\right)+2f\left(x_2\right)+f\left(x_3\right)\right]$

Where,

width="127" style="max-width: none; vertical-align: -26px;" $\begin{array}{rcl}f\left(x_0=0\right)& =& \sin \left(0^2\right)\\ & =& 0\end{array}$

And

width="168" style="max-width: none; vertical-align: -57px;" $\begin{array}{rcl}f\left(x_1=\frac{1}{6}\right)& =& \sin {\left(\frac{1}{6}\right)}^2\\ & =& \sin \left(\frac{1}{36}\right)\\ & \approx & 0.0004848\end{array}$

Again,

width="149" style="max-width: none; vertical-align: -68px;" $$\begin{gathered} f\left(x_1=\frac{1}{3}\right)=\sin {\left(\frac{1}{3}\right)}^2 \\ =\sin \left(\frac{1}{9}\right) \\ \approx 0.001939 \end{gathered}$$

Hence,

width="353" style="max-width: none; vertical-align: -44px;" $\begin{array}{rcl}I& \approx & \frac{1}{3}[0+4(0.0004848)+2(0.001939)+0.004363]\\ & \approx & \frac{1}{18}[0+0.001939+0.003878+0.004363]\end{array}$

 

Implies that width="82" style="max-width: none; vertical-align: -4px;" $I\approx 0.00057$

Part (c) Step 1: Given information

The integral is $I={\int }_0^{0.5}\sin \left(x^2\right)dx$

Part (c) Step 2: Calculation

Since the Maclaurin series for$\sin \left(x\right)$ is

$\sin x=\sum _{k=0}^{\infty }\frac{(-1{)}^k}{(2k+1)!}{x}^{2k+1}$

So, the Maclaurin series for $\sin \left(x^2\right)$ can be found by substituting x by $x^2$ in the Maclaurin series of $\sin x$

Therefore,

$\begin{array}{rcl}\sin \left(x^2\right)& =& \sum _{k=0}^{\infty }\frac{(-1{)}^k}{(2k+1)!}{\left(x^2\right)}^{2k+1}\\ & =& \sum _{k=0}^{\infty }\frac{(-1{)}^k}{(2k+1)!}{x}^{4k+2}\end{array}$

Implies that

$\sin \left(x^2\right)=\sum _{k=0}^{\infty }\frac{(-1{)}^k}{(2k+1)!}{x}^{4k+2}$

Part (d) Step 1: Given information

The integral is $I={\int }_0^{0.5}\sin \left(x^2\right)dx$

Part (d) Step 2: Calculation

Since $I={\int }_0^{05}\sin \left(x^2\right)dx$

Therefore,

$\begin{array}{rcl}I& =& {\int }_0^{a5}\left[\sum _{k=0}^{\infty }\frac{(-1{)}^k}{(2k+1)!}{x}^{4k+2}\right]dx\\ & =& \sum _{k=0}^{\infty }\frac{(-1{)}^k}{(2k+1)!}{\int }_0^{05}{x}^{4k+2}dx\\ & =& \sum _{k=0}^{\infty }\frac{(-1{)}^k}{(2k+1)!}{\left[\frac{x^{4k+3}}{4k+3}\right]}_0^{05}\\ & =& \sum _{k=0}^{\infty }\frac{(-1{)}^k}{(2k+1)!(4k+3)}(0.5{)}^{4k+3}\end{array}$

Now, to evaluate the definite integral of the series / from 0 to 0.5, let us find the sum of the terms that are greater than 0.001. The resultant term is the approximation.

Thus,

$\begin{array}{rcl}I& =& \frac{1}{3}(0.5{)}^3-\frac{1}{3!\left(7\right)}(0.5{)}^7+\frac{1}{5!\left(11\right)}(0.5{)}^{11}-\frac{1}{7!\left(15\right)}(0.5{)}^{15}+\frac{1}{9!\left(19\right)}(0.5{)}^{19}-\dots \\ & =& 0.041666-0.00018601+\dots \end{array}$

Since$0.00018601<<0.001$

$0.00018601<<0.001$

So, we consider first three terms to approximate I to within 0.001 of its actual value.

Therefore,

$I\approx 0.0416$