EBK NUMERICAL METHODS FOR ENGINEERS
EBK NUMERICAL METHODS FOR ENGINEERS
7th Edition
ISBN: 8220100254147
Author: Chapra
Publisher: MCG
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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.

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