Multiple-Choice Test Simpson’s 1/3 Rule Integration ...

[Pages:11]Multiple-Choice Test Simpson's 1/3 Rule Integration COMPLETE SOLUTION SET

1. The highest order of polynomial integrand for which Simpson's 1/3 rule of integration is exact is (A) first (B) second (C) third (D) fourth

Solution The correct answer is (C).

Simpson's 1/3 rule of integration is exact for integrating polynomials of third order or less.

Although Simpson's 1/3 rule is derived by approximating the integrand by a second order

polynomial, the area under the curve is exact for a third order polynomial. Without proof it can

be

shown that

the truncation

error

in Simpson's 1/3

rule

is

Et

=

- (b - a)5 2880

f (4) ( ),

a < < b.

Since the fourth derivative of a third order polynomial is zero, the truncation error would be zero.

Hence Simpson's 1/3 rule is exact for integrating polynomials of third order or less.

2.2

2. The value of e xdx by using 2-segment Simpson's 1/3 rule most nearly is

0.2

(A) 7.8036 (B) 7.8423 (C) 8.4433 (D) 10.246

Solution The correct answer is (B).

The multiple segment equation for Simpson's 1/3 rule is

b

a

f (x)dx

b-a 3n

f

(x0 )

+

4

n-1

f

i =1 i =odd

n-2

(xi ) + 2 f (xi ) +

i=2 i =even

f (xn )

Using two-segments gives

a = 0.2 b = 2.2 n=2 h= b-a

n = 2.2 - 0.2

2 =1 x0 = 0.2 x1 = x0 + h = 0.2 +1 = 1.2 x2 = 1.2 +1 = 2.2

 ( ) 2.2 e x dx

0.2

2.2 - 0.2 3?2

f

(0.2)

+

2-1

4f

i=1 i=odd

(xi )

+

2

2-2

f

i=2 i=even

xi

+ f (2.2)

=

2.2 3

- ?

0.2 2

f

(0.2)

+

4

1

i=1 i=odd

f

( xi

)

+

0

2

i=2 i=even

f

( xi

)

+

f (2.2)

= 2.2 - 0.2 [ f (0.2) + 4 f (1.2) + f (2.2)]

3? 2

[ ] = 0.33333 e0.2 + 4 ? e1.2 + e2.2

= 0.33333[23.527]

= 7.8423

2.2

3. The value of e xdx by using 4-segment Simpson's 1/3 rule most nearly is

0.2

(A) 7.8036

(B) 7.8062

(C) 7.8423

(D) 7.9655

Solution The correct answer is (B).

The multiple segment equation for Simpson's 1/3 rule is

b

a

f (x)dx

b-a 3n

f

( x0

)

+

n-1

4

i=1 i=odd

f

( xi

)

+

n-2

2f

i=2 i=even

(

xi

)

+

f (xn )

Using 4 segments gives a = 0.2

b = 2.2

n=4

h= b-a n

= 2.2 - 0.2 4

= 0.5

2.2 e x dx

0.2

2.2 - 0.2 3? 4

f

4-1

(x0 ) + 4 f

i=1 i=odd

4-2

(xi ) + 2 f

i=2 i=even

(xi ) +

f

(x4 )

=

2.2 3

- ?

0.2 4

f

( x0

)

+

3

4

i=1 i=odd

f

( xi

)

+

2

2f

i=2 i=even

(

xi

)

+

f (x4 )

=

2.2 - 0.2 [ f

3? 4

(x0 ) +

4( f

(x1) +

f

(x3 ))+

2( f

(x2 ))+

f

(x4 )]

So

f (x) = ex f (x0 ) = f (0.2) = e0.2 = 1.2214

f (x1) = f (0.2 + 0.5) = f (0.7) f (0.7) = e0.7 = 2.0138

f (x2 ) = f (0.7 + 0.5) = f (1.2) f (1.2) = e1.2 = 3.3201

f (x3 ) = f (1.2 + 0.5) = f (1.7) f (1.7) = e1.7 = 5.4739

f (x4 ) = f (2.2) = e2.2 = 9.0250

2.2 e x dx

0.2

2.2 - 0.2 [ f

3? 4

(x0 ) +

4( f

(x1 ) +

f

(x3 ))+ 2( f

(x2 ))+

f

(x4 )]

= 2.2 - 0.2 [1.2214 + 4(2.0138 + 5.4739)+ 2(3.3201)+ 9.0250]

3? 4

= 0.16667[1.2214 + 29.951+ 6.6402 + 9.0250]

= 7.8062

4. The velocity of a body is given by

v(t) = 2t,

1t 5

= 5t 2 + 3, 5 < t 14

where t is given in seconds, and v is given in m/s. Using two-segment Simpson's 1/3 rule, the distance in meters covered by the body from t = 2 to t = 9 seconds most

nearly is

(A) 949.33

(B) 1039.7

(C) 1200.5

(D) 1442.0

Solution The correct answer is (C).

The multiple segment equation for Simpson's 1/3 rule is

b

a

f (x)dx

b-a 3n

f

( x0

)

+

n-1

4

i=1 i=odd

f

( xi

)

+

n-2

2f

i=2 i=even

(

xi

)

+

f (xn )

a =2

b=9

n=2

h = b-a n

= 9-2 2

= 3.5

9

2

v(t)dt

9-2 3? 2

v(t0 )

+

2-1

4 v(ti )

i=1 i=odd

+

2-2

2 v(ti )

i=2 i=even

+

v(t2 )

=

9-2 3? 2

v(t0 )

+

1

4 v(ti )

i=1 i=odd

+

0

2 v(ti )

i=2 i=even

+

v(t2 )

=

9 3

- ?

2 2

[v(t0

)

+

4(v(t1

))

+

v(t2

)]

So

v(t) = 2t,

1t 5

= 5t 2 + 3, 5 < t 14

v(t0 ) = v(2) = 2 ? 2 = 4 m/s

v(t1) = v(2 + 3.5) = v(5.5) v(5.5) = 5? 5.52 + 3 = 154.25 m/s

v(t2 ) = v(9) = 5? 92 + 3 = 408 m/s

9

2

v(t )dt

9-2 3? 2

[v(t0 )

+

4? v(t1)+

v(t2 )]

= 9 - 2 [v(2) + 4? v(5.5)+ v(9)]

3? 2

= 1.1667[4 + 4?154.25 + 408]

= 1200.5 m

19

5. The value of f (x)dx by using 2-segment Simpson's 1/3 rule is estimated as

3

702.039. The estimate of the same integral using 4-segment Simpson's 1/3 rule most

nearly is

(A) 702.039 + 8 [2 f (7) - f (11) + 2 f (15)]

3

(B) 702.039 + 8 [2 f (7) - f (11) + 2 f (15)]

23

(C) 702.039 + 8 [2 f (7) + 2 f (15)]

3

Solution

(D) 702.039 + 8 [2 f (7)2 f (15)]

23

The correct answer is (B).

Using 2-segment Simpson's 1/3 rule gives

19

3

f

( x)dx

19 - 3 3? 2

f

2-1

(x0 ) + 4 f

i=1 i=odd

2-2

(xi ) + 2 f

i=2 i=even

(xi ) +

f

(x2 )

=

19 - 3 3? 2

f

(x0 ) +

1

4f

i=1 i=odd

(xi ) +

f

(x2 )

= 19 - 3 [ f

3? 2

(x0 ) + 4 f

(x1 ) +

f

(x2 )]

702.039 19 - 3 [ f (3) + 4 f (11)+ f (19)]

3? 2

Using 4-segment Simpson's 1/3 rule gives

19

3

f

( x)dx

19 - 3 3? 4

f

4-1

(x0 ) + 4 f

i=1 i=odd

4-2

(xi ) + 2 f

i=2 i=even

(xi ) +

f

(x4 )

=

19 - 3 3? 4

f

3

(x0 ) + 4 f

i=1 i=odd

2

(xi ) + 2 f

i=2 i=even

(xi ) +

f

(x4 )

=

19 - 3 [ f

3? 4

(x0 )

+

4(

f

(x1 ) +

f

(x3 ))+

2( f

(x2 ))+

f

(x4 )]

= 19 - 3 [ f (3) + 4( f (7)+ f (15))+ 2( f (11))+ f (19)]

3? 4

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