numerical method

(a)
v(t)=at^3+bt^2+ct+d
v(0)=0.4=>d=0.4
v(1)=1=>a+b+c=0.6
v(2)=2.6=>4a+2b+c=1.1
v(4)=11.2=>16a+4b+c=2.7
So 3a+b=0.5
5a+b=0.7
a=0.1,b=0.2.c=0.3
v(t)=0.1t^3+0.2t^2+0.3t+0.4
v(3)=5.8
v''(3)=0.6(3)+0.4=2.2
S(3)=0.1t^4/4+0.2t^3/3+0.3t^2/2+0.4t | (0,4)=14.6667
(b)
a+b+c=5.9
9a+3b+c=12.1
25a+5b+c=19.9
So 4a+b=3.1
12a+2b=7
b=2.3,a=0.2,c=3.4
A=0.2x^3/3+2.3x^2/2+3.4x | (1,5) =49.4667
(ii) f'(2)=2ax+b=2(0.2)(2)+2.3=3.1
(iii) f(2)=0.2(2^2)+2.3(2)+3.4=0.8+4.6+3.4=8.8
f''(2)=2a=0.4