maximum value

(b) 1/3

f(x)=x³/3-4x²+16x-64/3

maximum at x=5

[t³/3-4t²+16t]₄^x

(x³/3-4x²+16x)-(64/3-64+64)

there was i guess a slightly more simple way to loko at these problems

\int_{a}^{x}{f(t)dt}

will have to maximise this, then we simply have to find x where f(x)=0

so that is at x=4

so we have to see end points.. (0 or 5)

note that x=4 here is a point of minima..

0 will give a -ve value! why?