Q1. The highest order of polynomial integrand for which  Simpson’s 1/3 rule of integration is exact is

first

second
third
fourth

 Q2. The value of by using two-segment Simpson's 1/3 rule is most nearly

7.8306

7.8423

8.4433

10.246

Q3. The value of by using four-segment Simpson's 1/3 rule is most nearly

7.8036
7.8062

7.8423
7.9655

Q4. The velocity of a body is given by

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

949.33

1039.7

1200.5

1442.0

Q5. The value of   by using two-segment Simpson’s 1/3 rule is estimated as 702.039.  The estimate of the same integral using four-segment Simpson’s 1/3 rule most nearly is

702.39 + 8/3 [2f(7)-f(11)+2f(15)]

702.39/2 + 8/3 [2f(7)-f(11)+2f(15)]

702.39 + 8/3 [2f(7)+2f(15)]

702.39/2 + 8/3 [2f(7)+2f(15)]

 Q6. The following data of the velocity of a body is given as a function of time.

The best estimate of the distance in meters covered by the body from t=4 to t=15 using combined Simpson’s 1/3^rd rule and the trapezoidal rule would be

354.70

362.50

368.00

378.80

Complete Solution

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