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Can someone suggest a really hard calculus problem that can be solved with the knowledge of a high school student ? I would really like to work my brains on something interesting .

Thanks a lot !

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closed as too broad by anorton, Claude Leibovici, Magdiragdag, Sami Ben Romdhane, Davide Giraudo Apr 19 at 8:57

There are either too many possible answers, or good answers would be too long for this format. Please add details to narrow the answer set or to isolate an issue that can be answered in a few paragraphs.If this question can be reworded to fit the rules in the help center, please edit the question.

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I guess every book has $\int\sqrt{\tan x}dx$? –  Awesome Apr 18 at 19:39
    
Mine apparently doesn't ! –  Victor Apr 18 at 19:41
    
Its not elegant... just long and a bit intuitive in middle. I can say its "hard". –  Awesome Apr 18 at 19:42
    
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I'm sorry to be "that guy," but... Is there actually an "answer" to this question? This is highly subjective, and is not a great fit for the Q&A format of SE. That said, I'm sure that a list of hard problems would be pretty fun to have; I'm just not sure if this is the place on the web. (Voting to close.) –  anorton Apr 19 at 4:13
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7 Answers 7

up vote 4 down vote accepted

Here's another one to see how you understand calculus.

$f(x)\geqslant 0, \forall x \geqslant 0$

$f(x)\leqslant c\int_0^xf(t)dt, \forall x\geqslant 0 ,\exists c>0$

Prove that $f(x)$ is identically zero.

Solution : Let $F(x)=\int_0^xf(t)dt $

$$f(x)-cF(x)\leqslant 0$$

$$f(x)e^{-cx}-ce^{-cx}F(x)\leqslant0$$

$$(F(x)e^{-cx})'\leqslant 0$$

$$\text{Let }g(x)=F(x)e^{-cx}$$

$$\text{We know that } g(0)=0$$

$$\implies g(x)\leqslant0$$

$$F(x)e^{-cx}\leqslant0$$

$$\implies F(x)\leqslant 0$$

$$f(0)=0\implies f(x)\leqslant 0$$

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I must admit , I'm not completely sure what 'identical 0' means . Hope I haven't reached a level of honesty that borders on stupidity :D –  Victor Apr 18 at 20:02
    
@Victor Just f(x)=0 –  Awesome Apr 18 at 20:05
    
Cant we just pick x = 0 in the second condition to get $f(x) \le 0$ so $f(x) \equiv 0$ ? Or is there more to it? –  Sandeep Silwal Apr 18 at 20:13
    
@SandeepSilwal That would be $f(0)=0$ –  Awesome Apr 18 at 20:14
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Could you put the quantifiers in a more standard position? It's hard to tell whether you're saying "for all x, there exists a c such that..." or "there exists a c such that for all x, ...". –  user2357112 Apr 19 at 3:11
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This limit has been floating around since the 1990s (if not longer), though I haven't seen it "in the wild" since 1996:

If $a$ is real, evaluate

$$ \lim_{x\to\infty} e^{e^{e^{[x + e^{-(a + x + e^{x} + e^{e^{x}})}]}}} - e^{e^{e^{x}}}. $$

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This looks completely demonic ! –  Victor Apr 18 at 19:55
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The satan himself. –  Awesome Apr 18 at 19:56
    
Let met guess...0? No... –  Awesome Apr 18 at 19:56
    
1­­­­­­­­­­­­­­­? –  Awesome Apr 18 at 20:06
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$e^{-a}$??????? –  Awesome Apr 19 at 5:22
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This was on this years intervarsity paper in Ireland. It's kinda fun.

$\large{\int\limits_0^4 \frac{dx}{4+2^x}}$

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1/2 , tricky one , thanks ! –  Victor Apr 18 at 19:22
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Find the (real) value of $a$ such that the curves \begin{eqnarray} y &=& a^x \\ y &=& \log_a (x) \end{eqnarray} intersect exactly once. Find also the $x$ and $y$ values where they intersect. Note that's the logarithm of base $a$ in the second curve.

I think this is a pretty tough problem. It doesn't involve advanced calculus, but you need to know your exponential and logarithm functions and how to compute derivatives for such functions. But to my mind, the hard part is that it requires some clever reasoning. Have fun.

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The two functions ,namely a^x and log a (x) , have two intersection points : if 1<a<e^(1/e), they 'touch' each other at x=a if a=e^(1/e), and have no intersection points if a>e^(1/e) . –  Victor Apr 19 at 23:57
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This will test all of your analytical and mathematical skills.

$f(x)$ is a differentiable function and $g(x)$ is a double differentiable function such that $|f(x)|\leqslant 1$ and $f'(x)=g(x)$. If $$f(0)^2+g(0)^2=9$$

then prove that there exists some $c\in(-3,3)$ such that$ \ \ g(c) \cdot g''(c)<0$.

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Sorry to bother , but what does the dot betwen g(c) and g''(c)=0 exactly mean ? –  Victor Apr 18 at 19:20
    
Just multiplication. –  Awesome Apr 18 at 19:21
    
@Victor Just edited a typo. Please check again –  Awesome Apr 18 at 19:23
    
@Victor This one is really hard. –  Awesome Apr 18 at 19:26
    
Thanks a lot , this one does look really interesting ! –  Victor Apr 18 at 19:29
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If you haven't seen it before, then this should put your integration skills to the test:

$$\int \! \sec^3(x) \, \mathrm{d}x$$

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Easy one , but thanks , served good for warming up . This one took more time writing than thinking –  Victor Apr 18 at 19:03
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Try :

$$ \int e^{-2x}\left[(e^x)^2 +(x^x)^2\right]\ln x dx$$

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Let's see, $\int e^{-2x}\left[(e^x)^2 +(x^x)^2\right]\ln x dx$ $=$ $\int e^{-2x}\left[(e^2x) +(x^{2x})\right]\ln x dx$ $=$ $\int \left[1 +e^{-2x}(x^{2x})\right]\ln x dx$ $=$ $\int (\ln x +e^{-2x}x^{2x}\ln x )dx$ $=$ $\int \ln x dx + \int (e^{2x\ln x - 2x}\ln x )dx$. The second part looks interesting. $\int (e^{2x\ln x - 2x}\ln x )dx=\int(e^{2x(\ln x-1)}\ln x)dx=\int(e^{2x(\ln x-\ln e)}\ln x)dx=\int(e^{2x(\ln\dfrac xe)}\ln x)dx=\int(e^{(\ln({\dfrac xe}^{2x}))}\ln x)dx=\int({\dfrac xe}^{2x}\ln x)dx$. Now I really do have no idea where to go (we have $x^x$) –  Quincunx Apr 19 at 6:10
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@Quincunx I would start with $$\left(\frac x e\right) ^x =t$$ –  Awesome Apr 19 at 6:25
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