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

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 22, Problem 9P

(a)

To determine

To calculate: The given integral 21x(x+2)dx using numerical integration.

(a)

Expert Solution
Check Mark

Answer to Problem 9P

Solution: The value of 21x(x+2)dx using numerical integration is 0.34633.

Explanation of Solution

Given Information:

The given integral is,

21x(x+2)dx

Formula used:

Simpson’s 1/3 rule.

I1=h3[ y0+yn+4(y1+y3+...+yn1)+2(y2+y4+...+yn2) ]

Extended Midpoint rule for h=18.

I2=18(f(t0)+f(t1)+f(t2)+f(t3))

Calculation:

Calculate the analytical value of integral,

21x(x+2)dx =12[ ln(x)ln(x+2) ]2 =12([ ln()ln(+2) ][ ln(2)ln(2+2) ])=12([ 0 ][ 0.693 ])=0.3465

Rewrite the integral for finite intervals,

21x(x+2)dx =00.51t2 (t)1(1t+2)dt=00.511+2tdt

The function values are given in table below,

t 0.0625 0.125 0.1875 0.25 0.3125 0.375 0.4375 0.5
f(t) 0.941176 0.842105 0.761905 0.695652 0.64 0.592593 0.551724 0.516129

Apply numerical integration to simplify,

00.511+2t dt

00.511+2tdt =12(8)[ f(0.0625)+f(0.125)+f(0.1875)+f(0.25)+f(0.3125)+f(0.375)+f(0.4375)+f(0.5)]

Substitute the values from above table.

21x(x+2)dx =116[ 0.941176+0.842105+0.761905+0.695652+0.64+0.592593+0.551724+0.516129]=0.34633

Hence, the value of 21x(x+2)dx using numerical integration is 0.34633.

(b)

To determine

To calculate: The given integral 0eysin2y dy using numerical integration.

(b)

Expert Solution
Check Mark

Answer to Problem 9P

Solution: The value of 0eysin2y dy using numerical integration is 0.393523.

Explanation of Solution

Given Information:

The given integral is,

0eysin2y dy

Formula used:

Simpson’s 1/3 rule.

I1=h3[ y0+yn+4(y1+y3+...+yn1)+2(y2+y4+...+yn2) ]

Extended Midpoint rule for h=18.

I2=18(f(t0)+f(t1)+f(t2)+f(t3))

Calculation:

Calculate the analytical value of integral,

0eysin2y dy =[ ey(2sin2y+cos(2y)5)10 ]0 =110[ ey(2sin2y+cos(2y)5) ]0 =110(e(2sin2+cos(2)5))(e0(2sin20+cos(20)5))

Since e=0

. So integral is,

0eysin2y dy =110((0)(1(0+15)))=110((0)(1(4)))=110(4)=0.4

The function values are given in table below,

x 0 0.0625 0.125 0.1875 0.25 0.3125 0.375 0.4375 0.5
f(x) 0 0.048 0.139 0.219 0.26 0.258 0.222 0.168 0.112

Apply 4-application Simpson’s 1/3 rule for first part of integral,

I1=h3[ y0+yn+4(y1+y3+...+yn1)+2(y2+y4+...+yn2) ]

Here, y=f(x) and h=18

Substitute the values from above table,

I1=(20)0+4(0.048+0.219+0.258+0.168)+2(0.139+0.26+0.222)+0.11224 =0.344115

Rearrange the integral for calculation of second integral,

I2=0121t2 e1/tsin21tdt

The function values are changed for rearranged integral which is,

t 0 1 2 3
f(t) 0 0.0908 0.00142 0.303

Apply extended midpoint rule with h=18.

I2=18(f(t0)+f(t1)+f(t2)+f(t3))

Substitute the values from above table,

I2=18(0+0.0908+0.00142+0.303)=0.0494

The total integral is,

I=I1+I2

Substitute the value from above,

I=0.344115+0.0494=0.3935232

Hence, the value of 0eysin2y dy using numerical integration is 0.4.

(c)

To determine

To calculate: The given integral 01(1+y2)(1+y22)dy using numerical integration.

(c)

Expert Solution
Check Mark

Answer to Problem 9P

Solution: The value of 01(1+y2)(1+y22)dy using numerical integration is 0.919813.

Explanation of Solution

Given Information:

The given integral is,

01(1+y2)(1+y22)dy

Formula used:

Simpson’s 1/3 rule.

I1=h3[ y0+yn+4(y1+y3+...+yn1)+2(y2+y4+...+yn2) ]

Extended Midpoint rule for h=18.

I2=18(f(t0)+f(t1)+f(t2)+f(t3))

Calculation:

Calculate the analytical value of the integral,

01(1+y2)(1+y22)dy =01(y2+1)(2+y22)dy =02(y2+1)(2+y2)dy =201(y2+12)(y2+(2)2)dy

Recall the formula,

1y2+a2 dy=1atan1(ya)

Simplify further,

01(1+y2)(1+y22)dy =2[ tan1(y)12tan1(y2) ]0 =2[ (tan1()12tan1(2))(tan1(0)12tan1(02)) ]

Substitute tan1()=π2 and tan1(0)=0,

01(1+y2)(1+y22)dy =2[ (π212(π2))(012(0)) ]=2(π2π22)(00)=2×(π(21)22)=1.301252

Thus, the final value of integral is,

01(1+y2)(1+y22)dy =0.92

Hence, the value of 01(1+y2)(1+y22)dy analytically is 0.92.

The function values are given in table below,

x 0 0.0625 0.125 0.1875 0.25 0.3125 0.375 0.4375 0.5
f(x) 1 0.9127 0.711 0.4995 0.333 0.2191 0.1448 0.0972 0.0667

Apply 4-application Simpson’s 1/3 rule for first part of integral,

I1=h3[ y0+yn+4(y1+y3+...+yn1)+2(y2+y4+...+yn2) ]

Here, y=f(x) and h=18

Substitute the values from above table,

I1=(20)1+4(0.9127+0.4995+0.2191+0.0972)+2(0.711+0.3333+0.1448)+0.066724 =0.863262

Rearrange the integral for calculation of second integral,

I2=0121t2(1+1t2)(1+12t2) dt

The function values are changed for rearranged integral which is,

t 0 1 2 3
f(t) 0.007722 0.063462 0.148861 0.232361

Apply extended midpoint rule with h=18.

I2=18(f(t0)+f(t1)+f(t2)+f(t3))

Substitute the values from above table,

I2=18(0.007722+0.063462+0.148861+0.232361)=0.056551

The total integral is,

I=I1+I2

Substitute the value from above,

I=0.863262+0.056551=0.919813

Hence, the value of 01(1+y2)(1+y22)dy using numerical integration is 0.919813.

(d)

To determine

To calculate: The given integral 2yey dy using numerical integration.

(d)

Expert Solution
Check Mark

Answer to Problem 9P

Solution: The value of 2yey dy using numerical integration is 7.39695.

Explanation of Solution

Given Information:

The given integral is,

2yey dy

Formula used:

Simpson’s 1/3 rule.

I1=h3[ y0+yn+4(y1+y3+...+yn1)+2(y2+y4+...+yn2) ]

Extended Midpoint rule for h=18.

I2=18(f(t0)+f(t1)+f(t2)+f(t3))

Calculation:

Calculate the analytical value of the integral,

2yey dy

Apply Numerical integration to simplify,

2yey dy =[ (y1)ey ]2 =[ (1)e((2)1)e(2) ]=0(21)e2

Further simplify,

0(21)e2=e2 =7.39

Hence, the value of 2yey dy analytically is 7.39.

The function values are given in table below,

x 0 0.0625 0.125 0.1875 0.25 0.3125 0.375 0.4375 0.5
f(x) -14.78 -6.72 -2.72 0.824 0 0.303 0.368 0.335 0.2707

Apply 4-application Simpson’s 1/3 rule for first part of integral,

I1=h3[ y0+yn+4(y1+y3+...+yn1)+2(y2+y4+...+yn2) ]

Here, y=f(x) and h=18

Substitute the values from above table,

I1=(2(2))14.78+4(6.720.824+0.303+0.335)+2(2.72+0+0.368)+0.270724 =7.807

Rearrange the integral for calculation of second integral,

I2=0121t3e1t dt

The function values are changed for rearranged integral which is,

t 0 1 2 3
f(t) 0.000461 0.073241 0.1335696 1.214487

Apply extended midpoint rule with h=18.

I2=18(f(t0)+f(t1)+f(t2)+f(t3))

Substitute the values from above table,

I2=18(0.000461+0.0732418+0.1335696+1.214487)=0.410383

The total integral is,

I=I1+I2

Substitute the value from above,

I=7.80733+0.410383=7.39695

Hence, the value of 01(1+y2)(1+y22)dy using numerical integration is 7.39695.

(e)

To determine

To calculate: The given integral 012πex22 dx using numerical integration.

(e)

Expert Solution
Check Mark

Answer to Problem 9P

Solution: The value of 012πex22 dx using numerical integration is 0.499415.

Explanation of Solution

Given Information:

The given integral is,

012πex22 dx

Formula used:

Simpson’s 1/3 rule.

I1=h3[ y0+yn+4(y1+y3+...+yn1)+2(y2+y4+...+yn2) ]

Extended Midpoint rule for h=18.

I2=18(f(t0)+f(t1)+f(t2)+f(t3))

Calculation:

Calculate the given integral,

012πex22 dx

Rewrite the given integral,

012πex22 dx =12π0ex22 dx

Recall the formula,

0eax2 dx=12πa

Apply Analytical integration to simplify for a=12,

12π0ex22 dx =12π[ 12π12 ]=12π[ 122π ]=12 =0.5

Hence, the value of 012πex22 dx analytically is 0.5.

The function values are given in table below,

x 0 0.0625 0.125 0.1875 0.25 0.3125 0.375 0.4375 0.5
f(x) 0.399 0.387 0.352 0.301 0.242 0.183 0.130 0.086 0.054

Apply 4-application Simpson’s 1/3 rule for first part of integral,

I1=h3[ y0+yn+4(y1+y3+...+yn1)+2(y2+y4+...+yn2) ]

Here, y=f(x) and h=18

Substitute the values from above table,

I1=(20)0.399+4(0.387+0.301+0.183+0.086)+2(0.352+0.242+0.130)+0.05424 =0.47725

Rearrange the integral for calculation of second integral,

I2=01212π1t2e12t2 dt

The function values are changed for rearranged integral which is,

t 0 1 2 3
f(t) 0 0 0.024413063 0.152922154

Apply extended midpoint rule with h=18.

I2=18(f(t0)+f(t1)+f(t2)+f(t3))

Substitute the values from above table,

I2=18(0+0+0.024413063+0.152922154)=0.02217

The total integral is,

I=I1+I2

Substitute the value from above,

I=0.47725+0.02217=0.499415

Hence, the value of 012πex22 dx using numerical integration is 0.499415.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Your employer automatically puts 5 percent of your salary into a 401(k) retirement account each year. The account earns 10% interest. Suppose you just got the job, your starting salary is $60000, and you expect to receive a 2.5% raise each year. For simplicity, assume that interest earned and your raises are given as nominal rates and compound continuously. Find the value of your retirement account after 20 years
Compute the principal values of i¹² (i — 1)² and (i(i − 1))². - -
1 2 3 Consider the matrix A = 0 4 5. Give an example of 006 (a) a nonzero matrix B for which det(A + B) = det(A) + det(B); (b) a matrix C for which det(A+C)det(A) + det(C).
Knowledge Booster
Background pattern image
Advanced Math
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, advanced-math and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Text book image
Advanced Engineering Mathematics
Advanced Math
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Wiley, John & Sons, Incorporated
Text book image
Numerical Methods for Engineers
Advanced Math
ISBN:9780073397924
Author:Steven C. Chapra Dr., Raymond P. Canale
Publisher:McGraw-Hill Education
Text book image
Introductory Mathematics for Engineering Applicat...
Advanced Math
ISBN:9781118141809
Author:Nathan Klingbeil
Publisher:WILEY
Text book image
Mathematics For Machine Technology
Advanced Math
ISBN:9781337798310
Author:Peterson, John.
Publisher:Cengage Learning,
Text book image
Basic Technical Mathematics
Advanced Math
ISBN:9780134437705
Author:Washington
Publisher:PEARSON
Text book image
Topology
Advanced Math
ISBN:9780134689517
Author:Munkres, James R.
Publisher:Pearson,
Statistics 4.1 Point Estimators; Author: Dr. Jack L. Jackson II;https://www.youtube.com/watch?v=2MrI0J8XCEE;License: Standard YouTube License, CC-BY
Statistics 101: Point Estimators; Author: Brandon Foltz;https://www.youtube.com/watch?v=4v41z3HwLaM;License: Standard YouTube License, CC-BY
Central limit theorem; Author: 365 Data Science;https://www.youtube.com/watch?v=b5xQmk9veZ4;License: Standard YouTube License, CC-BY
Point Estimate Definition & Example; Author: Prof. Essa;https://www.youtube.com/watch?v=OTVwtvQmSn0;License: Standard Youtube License
Point Estimation; Author: Vamsidhar Ambatipudi;https://www.youtube.com/watch?v=flqhlM2bZWc;License: Standard Youtube License