2
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

So the area is maximized on the positive part area, the interval is then $[-1,2]$.

$\endgroup$
3
  • $\begingroup$ How did you deduce that the maximum area is on the positive area? $\endgroup$
    – idolo
    Nov 23 '17 at 8:01
  • $\begingroup$ @idolo Because negative numbers are less than positive ones. $\endgroup$
    – AmorFati
    Nov 23 '17 at 8:01
  • 1
    $\begingroup$ Just look at what happen if we put $[-1,3]$, the part $[2,3]$ contributes negative area, and eliminates part of area of $[-1,2]$. $\endgroup$
    – user284331
    Nov 23 '17 at 8:05
2
$\begingroup$

Well, in general you're trying to find:

$$ \begin{cases} \frac{\partial}{\partial\text{a}}\left\{\int_\text{a}^\text{b}\left(\text{n}_1\cdot x^2+\text{n}_2\cdot x+\text{n}_3\right)\right\}=-\text{n}_3-\text{a}\cdot\left(\text{n}_2+\text{n}_1\cdot\text{a}\right)=0\\ \\ \frac{\partial}{\partial\text{b}}\left\{\int_\text{a}^\text{b}\left(\text{n}_1\cdot x^2+\text{n}_2\cdot x+\text{n}_3\right)\right\}=\text{n}_3+\text{b}\cdot\left(\text{n}_2+\text{n}_1\cdot\text{b}\right)=0 \end{cases}\tag1 $$

$\endgroup$
2
$\begingroup$

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 $a=-1,b=2$.

$\endgroup$
1
$\begingroup$

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 $-1\le x\le2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.