Chapter 5, Problem 13P

To determine

To find:

The interval $\left[a,b\right]$ for which the value of the given integral is a maximum

Answer to Problem 13P

The interval of $\left[a,b\right]$ is $\left(-1,2\right)$ for which the value of the given interval is maximum.

Explanation of Solution

Given information:

The given integral is ${\displaystyle {\int }_a^b\left(2+x-x^2\right)dx}$

Define,

  $\begin{array}{l}f\left(a,b\right)={\displaystyle {\int }_a^b\left(2+x-x^2\right)dx}\\ ={\left[2x+\frac{x^2}{2}-\frac{x^3}{3}\right]}_a^b\\ =2b+\frac{b^2}{2}-\frac{b^3}{3}-2a-\frac{a^2}{2}+\frac{a^3}{3}\end{array}$

It is necessary to find $a<b$ for which the function $f\left(a,b\right)$ has the minimal value.

Now, the suspicious point

  $\begin{array}{l}{f}_a\left(a,b\right)=-2-a+a^2=0\\ f_b\left(a,b\right)=2+b-b^2=0\end{array}$

So the four points are obtain from the solution of this problem

  $\left(2,2\right),\left(2,-1\right),\left(-1,2\right),\left(-1,-1\right)$

Only the point $\left(-1,2\right)$ satisfies the condition $a<b$ .

  $f\left(-1,2\right)=4+2-\frac{8}{3}+2-\frac{1}{2}-\frac{1}{3}=4.5$

Note that the parabola $y=2+x-x^2$ has $"\cap "$ shape and intersects the $x$ -axis at the points $x=-1\text{ and }x=2$ which means that $\left(-1,2\right)$ is the maximal interval for which the function is not negative. Any changes of this interval will lead to adding negative areas or reducing positive areas.

Therefore, the interval of $\left[a,b\right]$ is $\left(-1,2\right)$ for which the value of the given interval is maximum.