Answer
$[a,b]=[-1,2]$
Work Step by Step
Recall:
(i) If $f(x)$ is above the $x-$axis for $a\leq x\leq b$, then $A=\int_a^bf(x)dx$ represents the area between the curve $y=f(x)$ and the $x-$axis between $x=a$ and $x=b$.
(ii) If $f(x)$ is under the $x-$axis for $a\leq x\leq b$, then $A=-\int_a^bf(x)dx$ represents area between the curve $y=f(x)$ and the $x-$axis between $x=a$ and $x=b$.
Given: $\int_a^b2+x-x^2 dx$
Consider $f(x)=2+x-x^2$ or $f(x)=(2-x)(1+x)$
This function describes a downward parabola that intercepts the $x-$axis at $x=-1$ and $x=2$.
To maximize the value of the integral, find the closed interval $[a,b]$ on which the parabola curve is above the $x-$axis.
That is,
$f(x)\geq 0$
$2+x-x^2\geq 0$
$x^2-x-2\leq 0$
$(x+1)(x-2)\leq 0$
$-1\leq x\leq 2$
Thus, $[a,b]=[-1,2]$.
