Chapter 6.5, Problem 15E

(a)

To determine

To calculate: The value of the integral using Simpson’s rule with n=4 approximation

Answer to Problem 15E

The value of integral is ${\displaystyle \underset{0}{\overset{2}{\int }}{x}^{3}dx}=4$

Explanation of Solution

Given information:

  ${\displaystyle {\int }_0^2{x}^{3}dx}$

Formula used:

  $S=\frac{h}{3}\left(y_0+4y_1+2y_2+4y_3+\mathrm{.....}+2y_{n-2}+4y_{n-1}+y_n\right)$

Calculation:

To approximate ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx$

  $S=\frac{h}{3}\left(y_0+4y_1+2y_2+4y_3+\mathrm{.....}+2y_{n-2}+4y_{n-1}+y_n\right)$

Where [ a , b ] is partitioned into an even number n of subintervals of equal length $h=\frac{b-a}{n}$

Now, apply the Simpson’s rule with

  $\begin{array}{l}n=4\\ f\left(x\right)=x^3\\ [a,b]=[0,2]\end{array}$

So,

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

  

Now, substitute the values from given table in this equation

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{2}{\int }}{x}^{3}dx=\frac{h}{3}\left(y_0+4y_1+2y_2+4y_3+y_4\right)}\\ =\frac{\frac{1}{2}}{3}\left(0+4\left(\frac{1}{8}\right)+2\left(1\right)+4\left(\frac{27}{8}\right)+8\right)\\ =\frac{1}{6}\left(0+\frac{1}{2}+2+\frac{27}{2}+8\right)\\ {\displaystyle \underset{0}{\overset{2}{\int }}{x}^{3}dx}=4\end{array}$

Conclusion:

The value of integral is ${\displaystyle \underset{0}{\overset{2}{\int }}{x}^{3}dx}=4$

(b)

To determine

To calculate: The exact value of the integral ${\displaystyle {\int }_0^2{x}^{3}dx}$

Answer to Problem 15E

The value of the integral is ${\displaystyle \underset{0}{\overset{2}{\int }}{x}^{2}dx=4}$

Explanation of Solution

Given information:

  ${\displaystyle {\int }_0^2{x}^{3}dx}$

Formula used:

  ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx=F\left(b\right)-F\left(a\right)$

Calculation:

If f is continuous at every point of [a, b], and if F is any anti-derivative of f on [a, b], then

  ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx=F\left(b\right)-F\left(a\right)$

This part of the fundamental theorem is also called the integral evaluation theorem.

Now, use the above formula to find the value of integral.

  $f\left(x\right)=x^3$

If F is any anti-derivative of f then

  $F\left(x\right)=\frac{x^4}{4}$

So,

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{2}{\int }}{x}^{3}dx=\left(\frac{2^4}{4}-\frac{0^4}{4}\right)}\\ =\frac{16}{4}-0\\ {\displaystyle \underset{0}{\overset{2}{\int }}{x}^{2}dx=4}\end{array}$

Conclusion:

The value of the integral is ${\displaystyle \underset{0}{\overset{2}{\int }}{x}^{2}dx=4}$