if `f(x) = underset(0)overset(x)int(t^2+2t+2)` dt where `x in [2,4]` then

Options

A. the minimum value of f(x) is `32/3`
B. the minimum value of f(x)Is 10
C. the maxium value of f(x) is 10
D. none of these

Answer

A

Explanation

We have
`f(x)=underset(0)overset(x)(t^2+2t+2)dt`
`rArr f(x)=x^2+2x+2=(x+1)^2+1 gt 0 ` for all x
`rArr` f(x) is strictly increasing on [ 2,4]
`therefore ` Maximum value of f(x)
`=f(4)=underset(0)overset(2)int(t^2+2t+2)dt = [(r^3)/3+t^2+2t]_0^4=136/3`
`=f(2)=underset(0)overeset(2)int(t^2+2t+2)dt=[(t^3)+t^2+2t]_0^2=136/3`