Chapter 9, Problem 64RE

a.

To determine

To identify: The value of $f\left(x\right)={\displaystyle \underset{0}{\overset{1}{\int }}{x}^2{e}^{x}dx}$ by using the Trapezoidal Rule with $n=2$

Answer to Problem 64RE

The required value is ${\displaystyle \underset{0}{\overset{1}{\int }}{x}^2{e}^{x}dx}=0.88566$

Explanation of Solution

Given information:

The given integral is $f\left(x\right)={\displaystyle \underset{0}{\overset{1}{\int }}{x}^2{e}^{x}dx}$ .

Consider the given integral.

By using the trapezoidal rule:

  $\begin{array}{c}{\displaystyle \underset{0}{\overset{1}{\int }}{x}^2{e}^{x}dx}=\frac{h}{2}\left[f\left(0\right)+2f\left(0.5\right)+f\left(1\right)\right]\\ =\frac{1}{4}\left[0+2\frac{e^{0.5}}{4}+e\right]\\ =0.88566\end{array}$ .

b.

To determine

To identify: The first three non-zero term of the Maclaurin series series for $x^2{e}^x$ to obtain the fourth-order Maclaurin polynomial $P_4\left(x\right)$ for $x^2{e}^x$ . Use ${\displaystyle {\int }_0^1{P}_4}(x)dx$ to obtain another estimate of ${\displaystyle {\int }_0^1{x}^2}{e}^{x}dx$ .

Answer to Problem 64RE

The first three non-zero terms of the series are $T_0=x^2,T_1=x^3,T_2=\frac{x^4}{2}$ .

The fourth order Maclaurin Polynomial is $P_4\left(x\right)=x^2+x^3+\frac{x^4}{2!}$ .

Another estimation is $0.68333$ .

Explanation of Solution

Given information:

The given integral is $f\left(x\right)={\displaystyle \underset{0}{\overset{1}{\int }}{x}^2{e}^{x}dx}$ .

Consider the given integral.

Maclaurin series is generated by function $e^x$ is given by:

  $e^x=1+x+\frac{x^2}{2!}+\mathrm{...}+\frac{x^n}{n!}+\mathrm{...}$ .

Multiply each side by $x^2$ .

  $\begin{array}{c}{x}^2{e}^x=x^2\left(1+x+\frac{x^2}{2!}+\mathrm{...}+\frac{x^n}{n!}+\mathrm{...}\right)\\ =x^2+x^3+\frac{x^4}{2!}+\mathrm{...}+\frac{x^{n+2}}{n!}+\mathrm{...}\end{array}$

Therefore, the first three non-zero terms of the series are $T_0=x^2,T_1=x^3,T_2=\frac{x^4}{2}$ .

Therefore, the fourth order Maclaurin Polynomial is $P_4\left(x\right)=x^2+x^3+\frac{x^4}{2!}$ .

Now,

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{1}{\int }}{P}_4\left(x\right)dx={\displaystyle \underset{0}{\overset{1}{\int }}\left(x^2+x^3+\frac{x^4}{2!}\right)dx}}\\ ={\left[\frac{x^3}{3}+\frac{x^4}{4}+\frac{x^5}{10}\right]}_0^1\\ =\frac{41}{60}\\ \approx 0.68333\end{array}$

c.

To determine

To identify: Why Trapezoidal Rule estimate obtained in part (a) is too large if the second derivative of $f\left(x\right)=x^2{e}^x$ is positive for all $x>0$ .

Answer to Problem 64RE

The function is concave up, the trapezoids used to estimate the area lie above the curve.

Explanation of Solution

Given information:

The given integral is $f\left(x\right)={\displaystyle \underset{0}{\overset{1}{\int }}{x}^2{e}^{x}dx}$ .

Consider the given integral.

Since the function is concave up, the trapezoids used to estimate the area lie above the curve.

Thus, the estimate is too large.

d.

To determine

To identify: Why Maclaurin series approximations to $f\left(x\right)$ for $x$ in $[0,1]$ will be too small if the second derivative of $f\left(x\right)=x^2{e}^x$ is positive for all $x>0$ .

Answer to Problem 64RE

All the derivative are positive and the remainder $R_n\left(x\right)$ , must be positive. This means that $P_n\left(x\right)$ must be smaller than $f\left(x\right)$ .

Explanation of Solution

Given information:

The given integral is $f\left(x\right)={\displaystyle \underset{0}{\overset{1}{\int }}{x}^2{e}^{x}dx}$ .

Consider the given integral.

Since all the derivative are positive and the remainder $R_n\left(x\right)$ , must be positive. This means that $P_n\left(x\right)$ must be smaller than $f\left(x\right)$ .

e.

To determine

To identify: The integration of ${\displaystyle {\int }_0^1{x}^2}{e}^{x}dx$ by using by parts method.

Answer to Problem 64RE

The required value is 0.71828.

Explanation of Solution

Given information:

The given integral is $f\left(x\right)={\displaystyle \underset{0}{\overset{1}{\int }}{x}^2{e}^{x}dx}$ .

Consider the given integral.

  ${\displaystyle {\int }_0^1{x}^2}{e}^{x}dx=x^2{e}^x-{\displaystyle \int 2x{e}^x}$

Further simplify.

  $\begin{array}{c}{\displaystyle {\int }_0^1{x}^2}{e}^{x}dx=x^2{e}^x-\left[2x{e}^x-{\displaystyle \int 2e^x}\right]\\ ={\left[x^2{e}^x-2x{e}^x+2e^x\right]}_0^1\\ =e-2\\ \approx 0.71828\end{array}$