Question

Find the approximations\({{\bf{T}}_{\bf{n}}}{\bf{,}}{{\bf{M}}_{\bf{n}}}\), and\({{\bf{S}}_{\bf{n}}}\)to the integral\(\int_{\bf{1}}^{\bf{4}} {\frac{{\bf{1}}}{{\sqrt x }}} {\bf{dx}}\)for \({\bf{n = 6}}\,{\bf{\& }}\,{\bf{12}}\) . Then compute the corresponding errors\({{\bf{E}}_{\bf{T}}}{\bf{,}}{{\bf{E}}_{\bf{M}}}\), and\({{\bf{E}}_{\bf{S}}}\). (Round your answers to six decimal places. You may wish to use the sum command on a computer algebra system.) What observations can you make? In particular, what happens to the errors when\({\bf{n}}\)is doubled?

Short answer

For\(n = 6\), \({E_T} = - 0.008966\),\({E_M} = 0.004428\),\({E_s} = - 0.000469\).

For\(n = 12\),\({E_T} = - 0.002269\),\({E_M} = 0.001131\),\({E_s} = - 0.000036\)

Step-by-step solution

Error in trapezoidal & midpoint rule:

Suppose\(\left| {{f^{\prime \prime }}(x)} \right| \le K\)for\(x \in (a,b)\).

If\({E_T}\)and\({E_M}\)are Errors in the Trapezoidal and Midpoint Rules, then\(\left| {{E_T}} \right| \le \frac{{K{{(b - a)}^3}}}{{12{n^2}}}\;\;\;{\rm{ and }}\;\;\;\left| {{E_M}} \right| \le \frac{{K{{(b - a)}^3}}}{{24{n^2}}}\).

Suppose thatfor. If\({{\bf{E}}_{\bf{S}}}\)is the error involved in using Simpson's Rule, then

Actual integral

The value of actual integral is:

\(\begin{aligned}{c}\int_1^4 {\frac{1}{{\sqrt x }}} dx = \left( {2\sqrt x } \right)_1^4\\ = 2\left( {\sqrt 4 - \sqrt 1 } \right)\\ = 2\left( {2 - 1} \right)\\ = 2\end{aligned}\)

Using Trapezoidal rule

The value of integral using Trapezoidal rule when\(n = 6\)is given by:

\(\begin{aligned}{c}\int_1^4 {\frac{1}{{\sqrt x }}} dx \approx \frac{{\frac{1}{2}}}{2}(f(1) + 2f(1.5) + 2f(2) + 2f(2.5) + 2f(3) + 2f(3.5) + f(4))\\ = \frac{3}{{10}}\left( {\frac{1}{{\sqrt 1 }} + \frac{2}{{\sqrt {1.5} }} + \frac{2}{{\sqrt 2 }} + \frac{2}{{\sqrt {2.5} }} + \frac{2}{{\sqrt 3 }} + \frac{2}{{\sqrt {3.5} }} + \frac{1}{{\sqrt 4 }}} \right)\end{aligned}\)

Therefore\({T_6} \approx 2.008966\)

When\(n = 12\):

\(\begin{aligned}{c}\int_1^4 {\frac{1}{{\sqrt x }}} dx \approx \frac{{\frac{1}{4}}}{2}(f(1) + 2f(1.25) + \ldots + 2f(3.75) + f(4))\\ = \frac{1}{8}\left( {\frac{1}{{\sqrt 1 }} + \frac{2}{{\sqrt {1.25} }} + \ldots + \frac{2}{{\sqrt {3.75} }} + \frac{1}{{\sqrt 4 }}} \right)\end{aligned}\)

Therefore\({T_{12}} \approx 2.002269\)

Using Midpoint  rule

The value of integral using Midpoint rule when\(n = 6\)is given by:

\(\begin{aligned}{c}\int_1^4 {\frac{1}{{\sqrt x }}} dx \approx \frac{1}{2}(f(1.25) + f(1.75) + f(2.25) + f(2.75) + f(3.25) + f(3.75))\\ = \frac{1}{2}\left( {\frac{1}{{\sqrt {1.25} }} + \frac{1}{{\sqrt {1.75} }} + \frac{1}{{\sqrt {2.25} }} + \frac{1}{{\sqrt {2.75} }} + \frac{1}{{\sqrt {3.25} }} + \frac{1}{{\sqrt {3.75} }}} \right)\end{aligned}\)

Therefore\({M_6} \approx 1.995572\)

When\(n = 12\):

\(\begin{aligned}{c}\int_1^4 {\frac{1}{{\sqrt x }}} dx \approx \frac{1}{4}(f(1.125) + f(1.375) + \ldots + f(3.625) + f(3.875))\\ = \frac{1}{4}\left( {\frac{1}{{\sqrt {1.125} }} + \frac{1}{{\sqrt {1.375} }} + \ldots + \frac{1}{{\sqrt {3.625} }} + \frac{1}{{\sqrt {3.875} }}} \right)\end{aligned}\)

Therefore\({M_{12}} \approx 1.998869\)

Using Simpson’s rule

The value of integral using Simpson’s rule when\(n = 6\)is given by:

\(\begin{aligned}{r}\int_1^4 {\frac{1}{{\sqrt x }}} dx \approx \frac{{\frac{1}{2}}}{3}(f(1) + 4f(1.5) + 2f(2) + 4f(2.5) + 2f(3) + 4f(3.5) + f(4))\\ = \frac{{\frac{1}{2}}}{3}\left( {\frac{1}{{\sqrt 1 }} + 4\frac{1}{{\sqrt {1.5} }} + 2\frac{1}{{\sqrt 2 }} + 4\frac{1}{{\sqrt {2.5} }} + 2\frac{1}{{\sqrt 3 }} + 4\frac{1}{{\sqrt {3.5} }} + \frac{1}{{\sqrt 4 }}} \right)\end{aligned}\)

Therefore\({S_6} \approx 2.000469\)

When\(n = 12\):

\(\begin{aligned}{c}\int_1^4 {\frac{1}{{\sqrt x }}} dx \approx \frac{{\frac{1}{3}}}{3}(f(1) + 4f(1.25) + 2f(1.5) + \ldots + 2f(3.5) + 4f(3.75) + f(4))\\ = \frac{1}{{12}}\left( {\frac{1}{{\sqrt 1 }} + 4\frac{1}{{\sqrt {1.25} }} + 2\frac{1}{{\sqrt {1.5} }} + \ldots . + 2\frac{1}{{\sqrt {3.5} }} + 4\frac{1}{{\sqrt {3.75} }} + \frac{1}{{\sqrt 4 }}} \right)\end{aligned}\)

Therefore\({S_{12}} \approx 2.000036\)

Corresponding Errors for\({\bf{n = 6}}\)

For Trapezoidal Rule:

\(\begin{aligned}{c}{E_T} = 2 - 2.008966\\ = - 0.008966\end{aligned}\)

For Midpoint Rule:

\(\begin{aligned}{c}{E_M} = 2 - 1.995572\\ = 0.004428\end{aligned}\)

For Simpson's Rule:

\(\begin{aligned}{c}{E_M} = 2 - 2.000469\\ = - 0.000469\end{aligned}\)

Corresponding Errors for \({\bf{n = 12}}\)

For Trapezoidal Rule:

\(\begin{aligned}{c}{E_T} = 2 - 2.002269\\ = - 0.002269\end{aligned}\)

For Midpoint Rule:

\(\begin{aligned}{c}{E_M} = 2 - 1.998869\\ = 0.001131\end{aligned}\)

For Simpson's Rule:

\(\begin{aligned}{c}{E_s} = 2 - 2.000036\\ = - 0.000036\end{aligned}\)

Observations

1. As\(n\) increases, all methods have more accurate approximations.

2. Errors in the Trapezoidal and Simpson's approximations have opposite signs to the Midpoint Rule.

3. The errors of the Trapezoidal and Midpoint rules decrease by a factor of approximately four when\(n\)is doubled. Whereas for Simpson's rule the error decreases by a factor of thirteen.

4. The Trapezoidal and Midpoint approximations are not as accurate as the Simpson's approximation.

5. The Trapezoidal error is approximately twice that of the Midpoint error and sixty times that of the Simpson's approximation.