Numerical Methods for Engineers
Numerical Methods for Engineers
7th Edition
ISBN: 9780073397924
Author: Steven C. Chapra Dr., Raymond P. Canale
Publisher: McGraw-Hill Education
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Chapter 22, Problem 1P

Use order of h 8 Romberg integration to evaluate

0 3 x e 2 x d x

Compare ε a and ε t .

Expert Solution & Answer
Check Mark
To determine

To calculate: The value of integral, 03xe2x dx using order of h8 Romberg integration and compare εa and εt.

Answer to Problem 1P

Solution:

The value of integral, 03xe2x dx is 504.69318.

Explanation of Solution

Given:

Integral, 03xe2x dx.

Formula used:

(1) Formula for h is given by,

h=ban

(2) Formula for true error percent is,

εt=| analytical value  I valueanalytical value |×100%

(3) Formula for relative percent error is,

εa=| I1,k  I2,k1I1,k |×100%

Calculation:

Consider the following integral:

I=03xe2x dx

This integral is a definite integral.

For analytical solution, simplify it as follows with respect to x, 03xe2x dx=[ xe2x2e2x4 ]03=[ 2xe2xe2x4 ]03=(2(3)e6e64)(0e0e04)=504.2859+0.25=504.5359

Hence, the analytical value of 03xe2x dx is 504.5359.

Let f(x)=xe2x

Use the application of the trapezoidal rule:

For n=1

, I=h2[ f(x0)+f(x1) ].

Here, a=0 and b=3.

Formula for h is given by,

h=ban

Substitute the value of a=0 and b=3 and simplify it as follows,

h=301=3

Values for x0 and x1 are,

x0=0x1=x0+h=0+3=3

Substitute the values in I

I=32[ f(0)+f(3) ].. .. .. (1)

Calculate f(0) as

f(0)=(0)e2(0)=0

Calculate f(3) as

f(3)=(3)e2(3)=3e6=1210.286

Substitute the values f(0)=0 and f(3)=1210.286 in equation (1) and simplify it as follows,

I=32[ 0+1210.286 ]=3×1210.2862=1815.429

For n=2

, I=h2[ f(x0)+2f(x1)+f(x2) ].

Formula for h is given by,

h=ban

Substitute the value of a=0, b=3 and n=2 and simplify it as follows,

h=302=1.5

Values for x0

, x1 and x2 are,

x0=0,x1=0+1.5=1.5x2=3

Substitute the values in I

I=1.52[ f(0)+2f(1.5)+f(3) ].. .. .. (2)

Calculate f(1.5) as

f(1.5)=(1.5)e2(1.5)=1.5e3=30.1283

Substitute the values f(0)=0, f(1.5)=30.1283 and f(3)=1210.286 in equation (2) and simplify it as follows,

I=1.52[ 0+2(30.1283)+1210.286 ]=1.5×1270.54262=952.90695

For n=3

, I=h2[ f(x0)+2f(x1)+2f(x2)+2f(x3)+f(x4) ].

Substitute the value of a=0, b=3 and n=4

and simplify it as follows,

h=304=0.75

Values for x0 and x1 are,

x0=0x1=0+0.75=0.75

Values for x2 and x3 are,

x2=0.75+0.75=1.5x3=1.5+0.75=2.25

And, x4=3

Substitute the values in I, I=0.752[ f(0)+2f(0.75)+2f(1.5)+2f(2.25)+f(3) ].. .. .. (3)

Calculate f(0.75) as

f(0.75)=(0.75)e2(0.75)=0.75e1.5=3.36127

Calculate f(2.25) as

f(2.25)=(2.25)e2(2.25)=2.25e4.5=202.5385

Substitute the values f(0)=0, f(1.5)=30.1283, f(0.75)=3.36127, f(2.25)=202.5385 and f(3)=1210.286 in equation (3) and simplify it as follows,

I=0.752[ 0+2(3.36127)+2(30.1283)+2(202.5385)+1210.286 ]=0.75×1682.342142=630.8783

For n=4

, I=h2[ f(x0)+2f(x1)+2f(x2)+2f(x3)+f(x4) ]

Formula for h is given by,

h=ban

Substitute the value of a=0, b=3 and n=4 and simplify it as follows,

h=308=0.375

Values for x0 and x1 are,

x0=0,x1=0+0.375=0.375

Values for x2 and x3 are,

x2=0.375+0.375=0.75x3=0.75+0.375=1.125

Values for x4 and x5 are,

x4=1.125+0.375=1.5x5=1.5+0.375=1.875

Values for x6 and x7 are,

x6=1.875+0.375=2.25x7=2.25+0.375=2.625

And,

x8=2.625+0.375=3

Substitute the values in I, I=0.752[ f(0)+2f(0.375)+2f(0.75)+2f(1.125)+2f(1.5)+2f(1.875)+2f(2.25)+2f(2.625)+f(3)].. .. .. (4)

Calculate f(0.375) as

f(0.375)=(0.375)e2(0.375)=0.375e0.75=0.793875

Calculate f(1.125) as

f(1.125)=(1.125)e2(1.125)=1.125e2.25=10.6737

Calculate f(1.875) as

f(1.875)=(1.875)e2(1.875)=1.875e3.75=79.72702

Calculate f(2.625) as

f(2.625)=(2.625)e2(2.625)=2.625e5.25=500.23645

Substitute the values obtained in equation (4) and simplify it as follows,

I=0.3752[ 0+2(0.793875)+2(3.36127)+2(10.6737)+2(30.1283)+2(79.72702)+2(202.5385)+2(500.23645)+1210.286]=0.375×2865.204232=537.22579

Use the Romberg integration method,

For O(h4)

with n=1,2

I43Im13Il

The Im

represents more accurate value and Il

represents the less accurate value.

Values for Im and Ii are,

Im=952.90695Il=1815.429

Substitute the values in I

and simplify it as follows,

I43(952.90695)13(1815.429)1270.5426605.143665.3996

For O(h4)

with n=2,4

Values for Im and Ii are,

Im=630.8783Il=952.90695

Substitute the values in I and simplify it as follows,

I43(630.8783)13(952.90695)841.171067317.63565523.535417

For O(h4)

with n=4,8

Values for Im and Ii are,

Im=537.22579Il=630.8783

Substitute the values in I

and simplify it as follows,

I43(537.22579)13(630.8783)716.30105210.29276506.00829

For O(h6)

with n=1,2

I1615Im115Il

Values for Im and Ii are,

Im=523.535417Il=665.3996

Substitute the values in I

and simplify it as follows,

I1615(523.535417)115(665.3996)558.43777844.359973514.077805

For O(h6)

with n=2,4

Values for Im and Ii are,

Im=506.00829Il=523.535417

Substitute the values in I

and simplify it as follows,

I1615(506.00829)115(523.535417)539.74217634.902361504.839815

For O(h8)

with n=4,8

Values for Im and Ii are,

Im=504.839815Il=514.077805

Substitute the values in I

and simplify it as follows,

I6463(504.839815)163(514.077805)512.8531458.159965504.69318

Formula for true error percent is,

εt=| analytical value  I valueanalytical value |×100%

Consider the following values,

analytical value=504.5359I value=1815.429

Substitute the values and simplify it as follows,

εt=| 504.53591815.429504.5359 |×100%=259.821%

Formula for true error percent for O(h4) is,

εt=| analytical value  I valueanalytical value |×100%

Consider the following values,

analytical value=504.5359I value=665.3996

Substitute the values and simplify it as follows,

εt=| 504.5359665.3996504.5359 |×100%=31.88%

Formula for true error percent for O(h6) is,

εt=| analytical value  I valueanalytical value |×100%

Consider the following values,

analytical value=504.5359I value=514.077805

Substitute the values and simplify it as follows,

εt=| 504.5359514.077805504.5359 |×100%=1.891%

Formula for true error percent for O(h8) is,

εt=| analytical value  I valueanalytical value |×100%

Consider the following values,

analytical value=504.5359I value=504.69318

Substitute the values and simplify it as follows,

εt=| 504.5359504.69318504.5359 |×100%=0.031%

Formula for relative error percent for O(h4) is,

εa=| I1,k  I2,k1I1,k |×100%

Here, the value of k=2 for O(h4), Now,

εa=| I1,2  I2,1I1,2 |×100%

Consider the following values,

I1,2=665.3996I2,1=952.90695

Substitute the values and simplify it as follows,

εa=| 665.3996952.90695665.3996 |×100%=43.208%

Formula for relative error percent for O(h6) is,

εa=| I1,k  I2,k1I1,k |×100%

Here, the value of k=3 for O(h6), Now,

εa=| I1,3  I2,2I1,3 |×100%

Consider the following values,

I1,3=514.077805I2,2=523.535417

Substitute the values and simplify it as follows,

εa=| 514.077805523.535417514.077805 |×100%=1.839%

Formula for relative error percent for O(h8) is,

εa=| I1,k  I2,k1I1,k |×100%

Here, the value of k=4 for O(h8), Now,

εa=| I1,4  I2,3I1,4 |×100%

Substitute the value of,

I1,4=504.69318I2,3=504.839815

Substitute the values and simplify it as follows,

εa=| 504.69318504.839815504.69318 |×100%=0.029%

Consider the following table that shows O(h4),O(h6),O(h8),εt,εa, Iteration1234εt259.82131.881.8910.031εa43.2081.8390.02911815.429665.3996514.077805504.693182952.90695523.535417504.8398154630.8783506.008298537.22579

Hence, the value of integral, 03xe2x dx is 504.69318.

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