Question

The difference between greatest and least value of function F(x) = ${\int }_0^x(t+1)dt$ on [2, 3] is 3.5. Prove it.

Answer 1

$F\left(x\right)={\int }_0^x(t+1)dt\Rightarrow F^{\prime }\left(x\right)=x+1>0,{\forall }_xϵ[2,3]$
Hence F(x) is increasing function in this interval.
Therefore greatest value is $F\left(3\right)={\int }_0^3(t+1)dt$
and least value $F\left(2\right)={\int }_0^2(t+1)dt$
hence required difference,
$h=F\left(3\right)-F\left(2\right)={\int }_0^3(t+1)dt-{\int }_0^2(t+1)dt$
$={\int }_2^3(t+1)dt=\frac{7}{2}$