Question

The difference between the greatest and the least value of the function f(x)=[integration with limits 0 to x] (t+1) dt on [2,3] is

Answer 1

F(x)=[integration with limits 0 to x] (t+1)dt
= [Limits 0 to x][(t^2)/2]+t

Applying the limits, we get
[(x^2)/2+x]-[(0^2)/2+0]
=[(x^2)/2]+x

So
F(x)=[(x^2)/2]+x

We need to find the difference between the greatest and the least value of the function in the interval[2 3].
The function reaches its maximum value at x=3
and minimum value at x=2

The difference between the greatest and the least value of the function F(X)=[(x^2)/2]+x
= F(3)-F(2)
F(3)={[(3^2)/2]+3} F(2)={[(2^2)/2]+2}
F(3)-F(2)
={(9/2)+3}-{(4/2)+2}

=(15/2)-(4)
=(7/2)=3.5