Finding $(a,b)$ that minimizes $I=\int_a^b(x^4−2x^2)\,\mathrm dx$

Question

Let $\displaystyle I=\int_a^b(x^4−2x^2)\,\mathrm dx$, then $I$ reaches the minimum when the ordered pair $(a,b)$ is:$$(-\sqrt2,0)\quad(0,\sqrt2)\quad(\sqrt2,-\sqrt2)\quad( -\sqrt2, \sqrt2)$$

I solved the integration and got $\dfrac{b^5}{5}-\dfrac{2b^3}{3}-\dfrac{a^5}{5}+\dfrac{2a^3}{3}$.

If I put $( -\sqrt2,0)$ or $ (0, \sqrt2)$, I get $\frac{(\sqrt2)^5}{5}-\frac{2(\sqrt2)^3}{3}$. Let this be $P$.

If I put $( \sqrt2, -\sqrt2)$, I get $-2P$. If I put $( -\sqrt2, \sqrt2)$, I get $2P$.

And since $P$ is negative, so the minimum is $2P$. So, the answer is $( -\sqrt2, \sqrt2)$. But this is quite a time consuming method. I wish if there was a quicker way to solve it (by observation or by graph or something, without so much calculation.)

Answer 1

$\def\d{\mathrm{d}}$Define $f(x) = x^4 - 2x^2$. Note that $\displaystyle I(a, b) = \int_a^b f(x) \,\d x$ is the the signed area enclosed by $x = a$, $y = 0$, $x = b$ and $y = f(x)$ if $a < b$, thus the signed area is minimized if $f(x) \leqslant 0$ for $x \in (a, b)$ and $f(x) \geqslant 0$ for $x \in \mathbb{R} \setminus (a, b)$. Since $f$ is positive on $(-∞, -\sqrt{2}) \cup (\sqrt{2}, +∞)$ and nonpositive on $[-\sqrt{2}, \sqrt{2}]$, then the integral $I(a, b)$ is minimized when $(a, b) = (-\sqrt{2}, \sqrt{2})$.

Answer 2

It's easy to sketch the graph of $$x^4-2x^2=x^2(x^2-2)=x^2(x-\sqrt2)(x+\sqrt2)$$

This is an even function, with a double root at $0$ which is a local maximum, and also roots of $-\sqrt2$ and $\sqrt2$. It goes to $\infty$ as $|x| \to \infty$.

We would pick $(-\sqrt{2}, \sqrt2)$ as it covers the whole negative region.

Checking that this is a better option compared to the other options:

The first option would only cover half of the region, it's the same for the second option though it covers another half.

The third solution would flip the region around and you will get a positive answer instead.

Remark: The global minimum for all real $a$ and $b$ indeed does not exists as we can let $a$ be arbitrarily large and $b$ be arbitrarily small but out of the $4$ options, the last option give the lowest value.

Answer 3

Simply check out the plot of the function that you are integrating, you will see that \begin{equation} (0, \sqrt{2}) \text { or }(-\sqrt{2}, 0) \end{equation} corresponds P which is negative so this is why the answer is: \begin{equation} (-\sqrt{2}, \sqrt{2}) \end{equation} Also, you need to check all the options because the minimum value of integration depends on the interval that is given and the plot is useful for this question because if you know the roots of the function and have a general idea of the behaviour you can guess the result of integration for this question you can simply see that minimum option is a negative integral which eliminates 2 options given.

Answer 4

@Ramit. Your mistake is to assume that $P>0$ which is false : $P=-\frac{8\sqrt{2}}{15}$.

As a consequence the minimum isn't $-2P$ but is $2P$ and so, at $(-\sqrt{2},\sqrt{2})$.