Chapter 10, Problem 64RE

(a)

To determine

The value is $f\left(x\right)=x^2{e}^x$

Answer to Problem 64RE

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

Explanation of Solution

Given information:

Consider the function,

  $f\left(x\right)=x^2{e}^x$

Formula used:

Trapezoidal rule is used.

Calculation:

Now, using trapezoidal rule

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

Conclusion:

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

(b)

To determine

The first three non-zero terms of $f\left(x\right)=x^2{e}^x$

Answer to Problem 64RE

The value is ${\displaystyle \underset{0}{\overset{1}{\int }}\left(x^2+x^3+\frac{x^4}{2!}\right)dx}=0.68333$

Explanation of Solution

Formula used:

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

Calculation:

Maclaurin series generated by function e^x is given by

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

Multiplying by x²,

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

Therefore, the first three non-zero terms of the series are $T_0=x^2{T}_1=x^3{T}_2={\frac{x}{2}}^4$

Therefore, fourth order Maclaurin polynomial is given by,

  $P_4\left(x\right)=x^2+x^3+\frac{x^4}{2!}$

Now,

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

Conclusion:

The value is ${\displaystyle \underset{0}{\overset{1}{\int }}\left(x^2+x^3+\frac{x^4}{2!}\right)dx}=0.68333$

(c)

To determine

To explain:The trapezoidal rule estimation in too large.

Answer to Problem 64RE

The estimate is too large.

Explanation of Solution

Given information:

The second derivative is $f(x)=x^2{e}^x$

Formula used:

The trapezoidal rule is used.

Calculation:

Since,

  $f\left(x\right)$ is concave up, the trapezoids used to estimate the area lie above the curve and thus the estimate is too large.

Conclusion:

The estimate is too large.

(d)

To determine

To explain:The Maclaurin series approximation for $f(x)=x^2{e}^x$ is too small.

Answer to Problem 64RE

The $P\left(x\right)$ must be smaller than $f\left(x\right)$

Explanation of Solution

Given information:

The second derivative is $f(x)=x^2{e}^x$

Formula used:

The maclaurin series approximation is used.

Calculation:

Since,

all the derivative are positive (and x > 0) the remainder, $R_n\left(x\right)$ must be positive.

This mean that $P\left(x\right)$ must be smaller than $f\left(x\right)$

Conclusion:

The $P\left(x\right)$ must be smaller than $f\left(x\right)$

(e)

To determine

The value of ${\displaystyle \underset{0}{\overset{1}{\int }}{x}^2{e}^x}dx$

Answer to Problem 64RE

The value of ${\displaystyle \underset{0}{\overset{1}{\int }}{x}^2{e}^x}dx=0.71828$

Explanation of Solution

Given information:

The given series,

  ${\displaystyle \underset{0}{\overset{1}{\int }}{x}^2{e}^x}dx$

Formula used:

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

Calculation:

Let $u=x^2\text{and}dv=e^{x}dx$

Therefore,

  $du=2xdx$ and $v=e^x$

Now,

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

Again, let $u=2x\text{and}dv=e^{x}dx$

Therefore,

  $du=2dx\text{and }v=e^x$

  $\begin{array}{l}{x}^2{e}^x-{\displaystyle \int 2x{e}^x}dx=x^2{e}^x-\left[2x{e}^x-{\displaystyle \int 2e^{x}dx}\right]\\ =x^2{e}^x-2x{e}^x+2e^x+C\\ x^2{e}^x-{\displaystyle \int 2x{e}^x}dx=\left(x^2-2x+2\right)e^x+C\end{array}$

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{1}{\int }}{x}^2{e}^x}dx=\left(x^2-2x+2\right)e^x{|}_0^1\\ =e-2\\ {\displaystyle \underset{0}{\overset{1}{\int }}{x}^2{e}^x}dx=0.71828\end{array}$

Conclusion:

The value of ${\displaystyle \underset{0}{\overset{1}{\int }}{x}^2{e}^x}dx=0.71828$