Question

The difference between the greatest and the least values of the function $f\left(x\right)={\int }_0^x(a{t}^2+1+\cos t)dt,a>0,x\in [2,3]$ is:

• A. $\frac{19}{3}a+1+\sin 3-\sin 2$
• B. $\frac{18}{3}a+1+2\sin 3$
• C. $\frac{18}{3}a-1+2\sin 3$
• D. none of these

Answer 1

The correct option is A $\frac{19}{3}a+1+\sin 3-\sin 2$

$f^{\prime }\left(x\right)=a{x}^2+1+\cos x$ is always +ive for $a>0,\forall xϵR$
Since
$\frac{dy}{dx}=+ive$
Hence $f\left(x\right)$ is an increasing function of $x$.
It will assume greatest value at $x=3$ and least value at $x=2$
Hence, the difference is: $f\left(3\right)-f\left(2\right)={\int }_2^3(a{t}^2+1+\cos t)dt$
$={[a\frac{t^3}{3}+t+\sin t]}_2^3=\frac{19}{3}a+1+(\sin 3-\sin 2)$