CHAPTER 71 NUMERICAL INTEGRATION

Transcription

1 CHAPTER 7 NUMERICAL INTEGRATION EXERCISE 8 Page 759. Evaluate using the trapezoidal rule, giving the answers correct to decimal places: + d (use 8 intervals) + = 8 d, width of interval = Hence, using the trapezoidal rule d + (.5) (. +.) =.5[.5559] =.569. Evaluate using the trapezoidal rule, giving the answers correct to decimal places: ln d (use 8 intervals) ln d, width of interval = = ln Hence, using the trapezoidal rule ln d (.5) ( ) =.5[7.968] = 6.979, John Bird

2 . Evaluate using the trapezoidal rule, giving the answers correct to decimal places: π / (sin θ) dθ (use 6 intervals) π / (sin θ) dθ, width of interval = π π = 6 8 or θ π/8 π/8 π/8 π/8 5π/8 6π/8 sinθ Hence, using the trapezoidal rule π / (sin θ) dθ ( +.96 ) = [.858] =.67. Evaluate using the trapezoidal rule, giving the answers correct to decimal places:. e d (use 7 intervals). e d, width of interval =. = e Hence, using the trapezoidal rule,. e d ( ) (.) = (.)[.5] =.8, John Bird

3 EXERCISE 8 Page 76. Evaluate using the mid-ordinate rule, giving the answers correct to decimal places: d + t t (use 8 intervals) d t, width of interval = =.5 + t 8 Hence, ordinates occur at,.5,.5,.75,.,.5,.5,.75 and., and mid-ordinates occur at.5,.75,.65,.875,.5,.75,.65 and.875 t t Hence, using the mid-ordinate rule, d + t t (.5)[ ] = (.5)[.96 ] =.. Evaluate using the mid-ordinate rule, giving the answers correct to decimal places: π / (use 6 intervals) + sinθ π / dθ, width of interval = + sinθ π π = rad or 5 6 Hence, ordinates occur at, 5,, 5, 6, 75 and 9, and mid-ordinates occur at 7.5,.5, 7.5, 5.5, 67.5 and 8.5. θ sinθ Hence, using the mid-ordinate rule, John Bird

4 + sinθ π / dθ [ ] = [.888] =.997. Evaluate using the mid-ordinate rule, giving the answers correct to decimal places: ln d (use intervals) ln d, width of interval = =. Hence, ordinates occur at.,.,.,.6,.8,.,.,.,.6,.8,., and mid-ordinates occur at.,.,.5,.7,.9,.,.,.5,.7 and ln Hence, using the mid-ordinate rule ln d (.)[ ] = (.)[.55] =.65. Evaluate using the mid-ordinate rule, giving the answers correct to decimal places: π / (cos ) d (use 6 intervals) π / (cos ) d, width of interval = π π = rad or 6 8 Hence, ordinates occur at,,,,, 5 and 6, and mid-ordinates occur at 5, 5, 5, 5, 5 and 55, John Bird

5 θ ( cos ) Hence, using the mid-ordinate rule, π / (cos ) d [ ] 8 = [.5768] =.799, John Bird

6 EXERCISE 8 Page 76. Evaluate using Simpson s rule, giving the answers correct to decimal places: π / (sin ) d (use 6 intervals) π / (sin ) d, width of interval = π π = rad or 5 6 π π π π 5π 6π (sin ) Hence, using Simpson s rule, π / (sin ) d ( +.) + ( ) + ( ) 6 = [ ] 6 = [.65] =.87. Evaluate using Simpson s rule, giving the answers correct to decimal places: +θ.6 dθ (use 8 intervals) +θ = 8.6 dθ, width of interval =.6. θ θ Hence, using Simpson s rule,.6 dθ +θ 5, John Bird

7 (.) (. +. ) + ( ) + ( ) (.) = [ + + ] = (.) [ 5.58 ] =.. Evaluate using Simpson s rule, giving the answers correct to decimal places:. sinθ dθ (use 8 intervals). θ. sinθ dθ, width of interval =.. =. (note that values of θ are in radians). θ 8 θ sinθ θ Hence, using Simpson s rule,.. ( ) ( ) sinθ dθ (.) θ + ( ) (.) = [ + + ] = (.) [.96 ] =.77. Evaluate using Simpson s rule, giving the answers correct to decimal places: π / cos d (use 6 intervals) π / cos d, width of interval = π π = rad or 5 6 π π π π 5π 6π cos , John Bird

8 Hence, using Simpson s rule, π / cos d ( + ) + ( ) + ( ) 6 = [ ] 6 = [ 6.5] = Evaluate using Simpson s rule, giving the answers correct to decimal places: π / e sin d (use intervals) π / e sin d, width of interval = π π = rad e sin π π π π 5π 6π 7π 8π 9π π Hence, using Simpson s rule, π / e sin d ( +.599) + ( ) + ( ) = [ ] 9 = [ 6.987] =.6 6. Evaluate using (a) integration, (b) the trapezoidal rule, (c) the mid-ordinate rule, (d) Simpson s rule. Give answers correct to decimal places. d (use 6 intervals) 7, John Bird

9 (a) d= d= = = 6 =.875 (b) Width of interval = = Hence, using the trapezoidal rule, d (.5) ( ) = (.5)[.85] =.7 (c) Mid-ordinates occur at.5,.75,.5,.75,.5 and Using the mid-ordinate rule, d (.5)[ ] = (.5)[.5] =.765 (d) Using the table of values from part (b), using Simpson s rule, d (. +.65) + ( ) + ( + ) (.5).5.8 (.5) = [ + + ] = (.5) [.967 ] = Evaluate using (a) integration, (b) the trapezoidal rule, (c) the mid-ordinate rule, (d) Simpson s rule. Give answers correct to decimal places. 6 d (use 8 intervals) ( ) 8, John Bird

10 (a) 6 du d Let u =, then ( ) d = and d = d u du u ( ) u Thus, d= = u du = = u = ( ) d = = =.585 ( ) 6 Hence, ( ) 6 (b) Width of interval = 6 = ( ) Hence, using the trapezoidal rule, 6 d ( ) (.5) ( ) = (.5)[.7595] =.588 (c) Mid-ordinates occur at.5,.75,.5,.75,.5,.75, 5.5 and ( ) Using the mid-ordinate rule, 6 d (.5)[ ( ) ] = (.5)[.656] =.58 (d) Using the table of values from part (b), using Simpson s rule, 6 d ( ) ( ) + ( ) + ( + + ) (.5) (.5) = [ + + ] 9, John Bird

11 = (.5) [ ] = Evaluate ( + ) d using (a) the trapezoidal rule, (b) the mid-ordinate rule, (c) Simpson s rule. Use 6 intervals in each case and give answers correct to decimal places. (a) Width of interval = = ( + ) Hence, using the trapezoidal rule, ( ) d + (.5) ( ) = (.5)[.875] =.9 (b) Mid-ordinates occur at.5,.75,.5,.75,.5,.75, 5.5 and ( + ) Using the mid-ordinate rule, ( + ) d (.5)[ ] = (.5)[.6] =.7 (c) Using the table of values from part (b), using Simpson s rule, ( ) d + ( ) + ( ) + ( + ) (.5).. (.5) = [ + + ] = (.5) [ 6. ] =.7, John Bird

12 9. Evaluate.7 d y using (a) the trapezoidal rule, (b) the mid-ordinate rule, (c) Simpson s. ( y) rule. Use 6 intervals in each case and give answers correct to decimal places. (a).7 d y then width of interval =.7. =.. 6 ( y ) y y ( ) Hence, using the trapezoidal rule, (.) d y. ( ) ( y ) = (.)[6.7675] =.677 (b) Mid-ordinates occur at.5,.5,.5,.5,.55 and.65 y ( y ) Using the mid-ordinate rule,.7 d y (.)[ ] ( y ). = (.)[6.78] =.67 (c) Using the table of values from part (a), using Simpson s rule, (.) d y ( + ) + ( + + ) + ( + ). ( y ) (.) = [ + + ] = (.) [.58 ] =.675, John Bird

13 . A vehicle starts from rest and its velocity is measured every second for 8 seconds, with values as follows: time t (s) velocity v (ms ) The distance travelled in 8. seconds is given by Estimate this distance using Simpson s rule, giving the answer correct to significant figures. 8. vd t 8. vdt. ( + 9.) + ( ) + ( ) ( )[ ] = (86.) = 8.8 m = [ ]. A pin moves along a straight guide so that its velocity v (m/s) when it is a distance (m) from the beginning of the guide at time t (s) is given in the table below. t(s) v(m/s) Use Simpson s rule with 8 intervals to determine the approimate total distance travelled by the pin in the. second period. Distance travelled by pin (.5) ( + ) + ( ) + ( ) [ ] = (.5) = (.5) [.98 ] =.85 m [ ], John Bird