Q.20 - Calculus

Question

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

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

(c) Use your answer from Exercise 19 (b) to 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 exists because the function $\sin \left(x^2\right)$ is continuous in the interval $\left[0,2\right]$.

(b) The actual value will be $0.05255$ by using the Simpson's method.

(c) Maclaurin seris and the integral power series can not be used.

Part (a) Step 1: The given information

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

Part (a) Step 2: Existence of the function

As the function $Sin\left(x^2\right)$ is continuous in the interval $\left[0,2\right]$.So, the integral $I$ exists.

Part (b) Step 1: Use of Simpson’s method

Let us apply the formula to approximate the integral I within $0.001$ of its real value using Simpson's approximation.

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

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

In the integral $a=0$, $b=2$ and $n=3$.

SO,

$$\begin{gathered} \Delta x=\frac{2-0}{3} \\ =\frac{2}{3} \end{gathered}$$

Here,

$\begin{array}{rcl}f\left(x_0=0\right)& =& \sin \left(0^2\right)\\ & =& 0\\ f\left(x_1=\frac{2}{3}\right)& =& \sin {\left(\frac{2}{3}\right)}^2\\ & =& \sin \left(\frac{4}{9}\right)\\ & \approx & 0.007757\\ & & \\ f\left(x_1=1\right)& =& \sin (1{)}^2\\ & =& \sin \left(1\right)\\ & \approx & 0.01745\\ & & \\ f\left(x_1=\frac{4}{3}\right)& =& \sin {\left(\frac{4}{3}\right)}^2\\ & =& \sin \left(\frac{16}{9}\right)\\ & \approx & 0.03102\\ f\left(x_1=2\right)& =& \sin {\left(2\right)}^2\\ & =& \sin \left(4\right)\\ & \approx & 0.06976\end{array}$

Hence,according to the value:

$\begin{array}{rcl}I& \approx & \frac{\Delta x}{3}\left[f\left(x_0\right)+4f\left(x_1\right)+2f\left(x_2\right)+4f\left(x_3\right)+f\left(x_4\right)\right]\\ I& \approx & \frac{\frac{2}{3}}{3}[0+4(0.007757)+2(0.01745)+4(0.03102)+0.06976]\\ & \approx & \frac{2}{9}[0+0.007756+0.0349+0.12408+0.06976]\\ I& \approx & 0.05255\end{array}$

In this way the  Simpson’s method can be used.

Part (c) Step 1: Use of two theorems

The maclaurin series of $\sin x$ is : $\sin x=\sum _{k=0}^{\infty }\frac{(-1{)}^k}{(2k+1)!}{x}^{2k+1}$

By using the series ,the value of  the function $\sin \left(x^2\right)$ is,

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

Apply the above value in the function:

$$\begin{gathered} I={\int }_0^2\sin \left(x^2\right)dx \\ I={\int }_0^2\left[\sum _{k=0}^{\infty }\frac{(-1{)}^k}{(2k+1)!}{x}^{4k+2}\right]dx \\ =\sum _{k=0}^x\frac{(-1{)}^k}{(2k+1)!}{\int }_0^2{x}^{4k+2}dx \\ =\sum _{k=0}^{\infty }\frac{(-1{)}^k}{(2k+1)!}{\left[\frac{x^{4\lambda +3}}{4k+3}\right]}_0^2 \\ =\sum _{k=0}^{\infty }\frac{(-1{)}^k}{(2k+1)!(4k+3)}(2{)}^{4k+3} \end{gathered}$$

Let's determine the sum of the terms that are higher than $0.001$ to calculate the definite integral of the series $/$ from $0$  to $0.5$. The approximation is the resultant term.

So,

$I=\frac{1}{3}(2{)}^3-\frac{1}{3!\left(7\right)}(2{)}^7+\frac{1}{5!\left(11\right)}(2{)}^{11}-\frac{1}{7!\left(15\right)}(2{)}^{13}+\frac{1}{9!\left(19\right)}(2{)}^{19}-\dots$

We can see that the term will keep increasing as we progress in our search for the approximation, implying that no term in the $/$ approximation will be less than $0.001$.

As a result, the Maclaurin series and the integral of $/$ cannot be used to approximate $I$.