Chapter 5.5, Problem 4QQ

a.

To determine

To calculate: The approximate value of $F\left(2\right)$ using the Trapezoidal Rule with four equal subdivisions of the closed interval $\left[0,2\right]$ if $F\left(x\right)={\displaystyle {\int }_0^x\sin \left(t^2\right)dt}$ for $0\le x\le 3$ .

Answer to Problem 4QQ

The resultant answer is $F\left(2\right)=0.744$ .

Explanation of Solution

Given information:

It is given that $F\left(x\right)={\displaystyle {\int }_0^x\sin \left(t^2\right)dt}$ for $0\le x\le 3$ .

Formula used: In the trapezoidal rule, if $\left[a,b\right]$ is portioned into n subintervals of equal length $h=\frac{b-a}{n}$ , the graph of $f\text{ on }\left[a,b\right]$ can be approximated by a straight line segment at each subinterval. The region between the curve and the x -axis is then approximately by the trapezoids, the area of each trapezoid being the length of its horizontal “altitude” times the average of two vertical “bases”.

That is ${\displaystyle {\int }_0^{b}f\left(x\right)dx}\approx \frac{h}{2}\left(y_0+y_1+y_2+\cdots +y_{n-1}+y_n\right)$ ,

where $y_0=f\left(a\right),y_1=f\left(x_1\right)\text{, }\cdots ,y_{n-1}=f\left(x_{n-1}\right)\text{, and }{y}_n=f\left(b\right)$ .

Calculation:

Consider the given function:

  $F\left(x\right)={\displaystyle {\int }_0^x\sin \left(t^2\right)dt}$

Now apply the trapezoid rule with $n=4,f\left(t\right)=\sin \left(t^2\right),\text{ and }\left[a,b\right]=\left[0,2\right]$ .

  $\begin{array}{c}h=\frac{b-a}{n}\\ =\frac{2-0}{4}\\ =\frac{1}{2}\end{array}$

Now find $F\left(2\right)={\displaystyle {\int }_0^x\sin \left(t^2\right)dt}$ .

  $\begin{array}{c}F\left(2\right)={\displaystyle {\int }_0^x\sin \left(t^2\right)dt}\\ =\frac{\frac{1}{2}}{2}\left[\sin 0+2\sin \left({0.5}^2\right)+2\sin \left(1^2\right)+2\sin \left({1.5}^2\right)+2\sin \left(2^2\right)\right]\\ =0.744\end{array}$

Therefore, the value is $F\left(2\right)=0.744$ .

b.

To determine

To state: The interval or intervals at which $F$ is increasing.

Answer to Problem 4QQ

The resultant answer is $\left[0,\sqrt{\pi }\right]\text{ and }\left[2\sqrt{\pi },3\right]$ .

Explanation of Solution

Given information:

It is given that $F\left(x\right)={\displaystyle {\int }_0^x\sin \left(t^2\right)dt}$ for $0\le x\le 3$ .

Consider the function $F\left(x\right)={\displaystyle {\int }_0^x\sin \left(t^2\right)dt}$ ,

  $\begin{array}{c}{F}^{\text{'}}\left(x\right)=\frac{d}{dx}{\displaystyle {\int }_0^x\sin \left(t^2\right)dt}\\ =\sin \left(x^2\right)\end{array}$

Then $F$ is increasing if $F^{\text{'}}\left(x\right)>0\text{ on }\left[0,\sqrt{\pi }\right]\text{ and }\left[2\sqrt{\pi },3\right]$ .

Similarly, $F$ is decreasing if $F^{\text{'}}\left(x\right)<0\text{ on }\left[\sqrt{\pi },2\sqrt{\pi }\right]$ .

Therefore, $F$ is increasing on intervals $\left[0,\sqrt{\pi }\right]\text{ and }\left[2\sqrt{\pi },3\right]$ for $F^{\text{'}}\left(x\right)>0$ .

c.

To determine

To state: The value of ${\displaystyle {\int }_0^x\sin \left(t^2\right)dt}$ in terms of k if the average rate of charge of $F$ on the closed interval $\left[0,3\right]$ is k .

Answer to Problem 4QQ

The resultant answer is ${\displaystyle {\int }_0^x\sin \left(t^2\right)dt}=3k$ .

Explanation of Solution

Given information:

The given average rate of charge of $F$ on the closed interval $\left[0,3\right]$ is k .

Consider the function $F\left(x\right)={\displaystyle {\int }_0^x\sin \left(t^2\right)dt}$ ,

Now find $F^{\text{'}}\left(x\right)=\frac{d}{dx}{\displaystyle {\int }_0^x\sin \left(t^2\right)dt}$ .

  $\begin{array}{c}{F}^{\text{'}}\left(x\right)=\frac{d}{dx}{\displaystyle {\int }_0^x\sin \left(t^2\right)dt}\\ =\sin \left(x^2\right)\end{array}$

Similarly, $F^{\text{'}}\left(t\right)=\sin \left(t^2\right)\text{ and }k=\sin \left(t^2\right)$ .

Now find ${\displaystyle {\int }_0^3\sin \left(t^2\right)dt}$ .

  $\begin{array}{c}{\displaystyle {\int }_0^3\sin \left(t^2\right)dt}={\displaystyle {\int }_0^{3}kdt}\\ =3k-3\left(0\right)\\ =3k\end{array}$

Therefore, the value of ${\displaystyle {\int }_0^x\sin \left(t^2\right)dt}$ in terms of $k\text{ is }3k$ .