a)
V=5t2+2t−t24
distance=∫25(5t2+2t−t24)dt
=[35t3+t2+t4]52
=35(5)3+(5)2+54−(35(2)3+(2)2+24)
=214.8(units)
b) The Simpson's 1/3 rule
Approximate the integral ∫25(5t2+2t−t24)dt with n=6 using the Simpson's rule.
∫abV(t)dt=3Δt(V(t0)+4V(t1)+2V(t2)+...
+2V(tn−2)+4V(tn−1)+V(tn)) We have that V(t)=5t2+2t−t24,a=2,b=5,n=6
Then Δt=nb−a=65−2=0.5
Divide the interval [2,5] into n=6 subintervals of the length Δt=0.5 with the following endpoints: a=2,2.5,3,3.5,4,4.5,5=b.
Now, just evaluate the function at these endpoints.
V(t0)=V(2)=23
4V(t1)=4V(2.5)=142.44
2V(t2)=2V(3)=9910
4V(t3)=4V(3.5)=4913313
2V(t4)=2V(4)=175.5
4V(t5)=4V(4.5)=8135657
V(t6)=V(5)=134.84
61(23+142.44+9910+4913313+175.5
+8135657+134.84)≈214.7991(units) Theanswer for distance from part A 214.8 units and from part B 214.799 units are very close
214.8∣214.799−214.8∣×100%=0.0005%
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