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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 found $f'(x)$ but it is not very straight-forward on how to determine the exact nature of the function. My idea to compare the area under curve rather than explicitly evaluate the integral.

How to go about it?

A similar question is here but there are no solutions.

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You remember the graph of $\tan x$? The one that is flat near zero but rapidly zooms off to infinity around $\frac\pi2 \approx 1.57$? Maybe the area under the curve around $x=0$ will be less than the area under curve around the region $x=1$!

Hint upon hint.

Notice also that $f(x)$ is in terms of positive $\tan x$ only, so we don't have to fanangle too much.

Lastly, notice that the bounds of the integrals have a difference of 1.

This should allow you to reason out the answer with out any messy number. Only visualization.

Comment any questions, and have fun mathing!

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  • $\begingroup$ I dont quite understand this statement of yours :"Maybe the area under the curve around $x=0$ will be less than the area under curve around the region $x=1$!" . Specifically, what do you mean when using the word "around" ? Do you mean some neighborhood about the point ? If so, this becomes much unreasonable, as there seems to be no reason to observe neighborhoods. I think this needs a circumstantial clarification! $\endgroup$
    – Arthur
    Commented Apr 28, 2023 at 14:32
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    $\begingroup$ @FdstZfsy Looking back at my answer, I'd say I am very wrong in my train of thought, and could've phrased myself better. My point with neighborhoods was the area of a region around x=0 would be smaller than around x=1, but it is irrelevant for the question because I didn't account for the -3tan(x) term. I wonder if a better approach might be to define a function $I(n) = \int_n^{n+1}$, take the derivative, see if it is increasing or decreasing, and apply it logically to the given function. Thank you for pointing out my error though, I appreciate it! $\endgroup$
    – Christopher Marley
    Commented Apr 29, 2023 at 7:01

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