Question

Let $f(a,b)={\int }_a^b(x^2-4x+3)dx,(b>a)$ then

• A. $f(a,3)$ is least when $a=1$
• B. $f(4,b)$ is an increasing function $\forall b\ge 4$
• C. $f(0,b)$ is least for $b=2$
• D. $min\left\{f(a,b)\right\}=-\frac{4}{3}\forall a,b\in R$

Answer 1

The correct option is A $f(a,3)$ is least when $a=1$

First of all integrate the function.

${\int }^{}(x^2-4x+3)dx=[\frac{x^3}{3}-4\frac{x^2}{2}+3x]$

Apply the limits,

$f(a,b)={[\frac{x^3}{3}-4\frac{x^2}{2}+3x]}_a^b$

Here $b=3$

Therefore,

$f(a,3)={[\frac{x^3}{3}-4\frac{x^2}{2}+3x]}_a^b=[\frac{27}{3}-2\times 9+9]-[\frac{a^3}{3}-2a^2+3a]=2a^2-\frac{a^3}{3}-3a$

Hence,

Differentiate the above function to find the maximum or minimum.

$f^{\prime }(a,3)=4a-a^2-3=0$

therefore $a=1,3$

Check whether the function is minimum or maximum.

$f^{\prime \prime }(a,3)=4-2a=2$ which is greater than $0$.

Hence the function $f(a,3)$ is minimum at $a=1$.