Calculus – Comparing Integrals of f(x)=(tan(x))^(3/2)-3tan(x)+(tan(x))^(1/2)

calculusintegration

Let $f(x)=(\tan x)^\frac32-3\tan x+\sqrt{\tan x}$. Consider the integrals $$I_1=\int_0^1f(x)dx$$
$$I_2=\int_{0.3}^{1.3}f(x)dx$$
$$I_3=\int_{0.5}^{1.5}f(x)dx$$
Then, prove that $I_1>I_3>I_2$

I tried out a lot of things. This question has been repeated twice at this site but no proper solution was found. Some found solutions using graphs. However, This is a question from a high school exam that does not permit the use of graphing softwares.

Here are a few of my approaches that failed:

  1. Creating a function $g(x)$ such that:$$ g(x)=\int_{x}^{x+1} f(t)dt$$ and differentiate it using Newton-Leibnitz. Nothing came out.It gives $g(x)=f(x+1)-f(x)$. The derivative of $f(x)$ does not reveal much to comment about $f(x+1)$ and $f(x)$.
  2. Integrating the expression and finding out the primitive. I tried substituting $t=\sqrt{\tan x}$ and proceeding further using algebraic twins. I did find a primitive and here it is:
    $$2(\ln(\frac{\sqrt{\tan x}+\sqrt{\cot x}-\sqrt{2}}{\sqrt{\tan x}+\sqrt{\cot x}+\sqrt{2}})+ \sqrt{\tan x})-3\ln{\sec x}$$
    Go ahead and put the limits and compare them if you dare. I won't stop you. If you find something out,please do tell me.
  3. Graphing by converting it into a reduced cubic and seeing if one or more areas are comparable enough.(It is a hand graph, so no use).But I got to see something interesting. I found out that the roots of the reduced cubic are $\frac{3+\sqrt{5}}{2}$ and $\frac{3-\sqrt{5}}{2}$ which can be written as $4cos^2 \frac{\pi}{10}$ and its reciprocal. But in the end the, it did not help me.

This is a problem from Bstat paper of ISI $2009$ and this problem has got me drunk.

Best Answer

We are only interested in the domain $0 < x < \frac{\pi}{2}$ since this contains all the domains of integration.

$$f(x)= \tan x \left ( \sqrt{\tan x} + \dfrac{1}{\sqrt{\tan x}} - 3\right ) = \tan x \times g(x)$$

We can sketch $g(x)$ easily by observing the following:

  • $g(x)$ is symmetric about $x=\frac{\pi}{4}$ since $f(\frac{\pi}{2} - x) = f(x)$
  • $g(x) \geq -1$ by AM/GM
  • $g(x)=0$ at $x \approx 0.145$ and $x \approx 1.426$. I am assuming you have access to a calculator that can compute up to square roots.

Sketch $g(x)$ and multiply ordinates with the graph of $\tan x$ to obtain a sketch of $f(x)$ that is sufficiently detailed to answer the question, with a few reasonable assumptions. The reliability of the answer depends on the level of care taken when multiplying ordinates.

$\displaystyle I_2 = \int_{0.3}^{1.3} f(x) \, dx$ is clearly the most negative since the region is purely below the $x$-axis and $x=1.3$ is reasonably close to $x=1.426$ so that it may capture a significant portion of the `negative region'. This narrows it down to either $I_1 > I_3 > I_2$ or $I_3 > I_1 > I_2$.

$\displaystyle I_1 = \int_{0}^{1} f(x) \, dx$ is a portion of the negative region that is offset by a tiny positive region from $0<x<0.145$.

$\displaystyle I_3 = \int_{0.5}^{1.5} f(x) \, dx$ is a larger portion of the negative region that is offset by a larger positive region in $1.426<x<1.5$. We can reasonably guess that as we are working close to the blow-up value of $\frac{\pi}{2} \approx 1.57$, the graph is steep at $x=1.426$ and the amount of positive region in $1.426 < x < 1.5$ is nullified by the amount of negative region at least the same interval backwards from $x=1.426$, but probably more since the positive region is nearer to the blow up point. So the net result is the negative region around $0.5<x<1.2$-ish.

Based off this assumption and by inspecting the graph, we can reasonably guess that $I_1 > I_3$ and hence the result.

Side Note: I can appreciate that you want a purely analytical solution that proves the direction without any doubt. I feel that is unreasonable taking into consideration that this is a multiple choice problem and not a written response problem. Why? Because in multiple choice problems where speed is important, candidates are expected to occasionally make educated guesses without necessarily knowing all the details. Examiners are conscious of this, and sometimes craft problems that are designed to

  • be punishing to candidates who mis-manage time by trying to properly find all the exact details rather than making a reasonably educated guess.
  • force the candidate to make educated assumptions because a purely analytical solution is impractical to find, or even impossible to find.

I know this is not the slick answer that you want. Perhaps there does exist a very slick way of properly determining the order. But if I were a sitting candidate with roughly thirty possibly equally painful other multiple choice problems to answer, this is exactly what I would do to.

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