Find the interval of $[a, b]$ to obtain a maximum integral. Find the interval $[a, b]$ for which the value of the integral 
$$\int_a^b (2 + x − x^2) dx$$
is a maximum. 
My Approach:
I've considered to let $f(x) = 2 + x - x^2 \implies f(x) = (2-x)(1+x)$. I've also considered that this is a negative quadratic graph, which shows that the positive area under the curve is bounded by the interval $-1$ to $2$. How do i then continue to maximise the value of the integral from the given interval $[a,b]$? 
Please explain as well, thanks. 
 A: Define $F(a,b)=\int_a^b f(x)~dx$ and you are looking for the global maximum of $F$ with the restriction $a\leq b$. You can do it by computing 
$$
\nabla F(a,b)=\begin{pmatrix}f(b)\\-f(a)\end{pmatrix}.
$$
and consider first the case $a<b$. So you have just $(-1,2)$ as a critical point.
Further, the Hessian matrix of $F$ at $(-1,2)$
$$
H_f(-1,2)=\begin{pmatrix}-3&0\\0&-3\end{pmatrix}
$$
yields, that $(-1,2)$ is a local maximum of $F$.
Since $F(a,a)=0$ and $F(-1,2)>0$ you can drop the boundary $a=b$.
Finally you can check that $F(a,b)\to c\in [-\infty,0]$ for $\|(a,b)\|\to \infty$ (with the restriction $a<b$) while $F(-1,2)>0$. From this you can deduce that $(-1,2)$ has to be the global maximum of $F$.
For the final step you also can show that $F(a,b)<0$ if $a>2$ or $b<-1$. Hence $F$ is nonpositive outside of a bounded area and the positive local maximum has to be the global maximum.
A: So the area is maximized on the positive part area, the interval is then $[-1,2]$.
A: Well, in general you're trying to find:
$$
\begin{cases}
\frac{\partial}{\partial\text{a}}\left\{\int_\text{a}^\text{b}\left(\text{n}_1\cdot x^2+\text{n}_2\cdot x+\text{n}_3\right)\right\}=-\text{n}_3-\text{a}\cdot\left(\text{n}_2+\text{n}_1\cdot\text{a}\right)=0\\
\\
\frac{\partial}{\partial\text{b}}\left\{\int_\text{a}^\text{b}\left(\text{n}_1\cdot x^2+\text{n}_2\cdot x+\text{n}_3\right)\right\}=\text{n}_3+\text{b}\cdot\left(\text{n}_2+\text{n}_1\cdot\text{b}\right)=0
\end{cases}\tag1
$$
A: You need the values of $a$ and $b$ such that the following is maximum $$\int_a^b (2 + x − x^2) dx.$$ Consider the function $$g(a,b)=\int_a^b (2 + x − x^2) dx=2(b-a)+\dfrac{b^2}{2}-\dfrac{a^2}{2}+\dfrac{a^3}{3}-\dfrac{b^3}{3}$$.Then taking partial derivatives with respect to $a$ and $b$ gives 
$$g_a(a,b)=-2-a+a^2=(a+1)(a-2)$$ 
$$g_b(a,b)=2+b-b^2=-(b+1)(b-2)$$.
Take double derivatives of $g$ with respect to $a$ and $b$ to check the maxima criterion and get the values $a=-1,b=2$.
A: Check the graph of $f(x)=2+x-x^2$. For a continuous function, you can view its integration as the area under the curve. So in this case as @user284331 wrote, the area is maximized for $-1\le x\le2$.
