Rigorous Calculations

Calculus Level 2

1 1 tan x 1 + x 2 + x 4 d x = ? \large \int_{-1}^{1} \dfrac{\tan x}{1+x^2+x^4} dx = \ ?


The answer is 0.000.

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1 solution

Nihar Mahajan
Jan 28, 2016

We use the property that if f f is an odd function then a a f ( x ) d x = 0 a f ( x ) d x + 0 a f ( x ) d x = 0 \int_{-a}^{a} f(x) dx=-\int_{ 0}^{a} f(x) dx +\int_{ 0}^{a} f(x) dx=0 . Note that f ( x ) = tan x 1 + x 2 + x 4 f(x)=\frac{\tan x}{1+x^2+x^4} satisfies f ( x ) = f ( x ) f(-x)=-f(x) , hence its an odd function and the result follows.

Congrats dude! Now let's start the integration contest! (After boards though )

Harsh Shrivastava - 5 years, 4 months ago

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U don't have any authority! :P let the contest be once a year (December). Do one thing keep a junior integration contest with an age limit.

Aditya Kumar - 5 years, 4 months ago

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That's what I meant :P

Any thoughts @Nihar Mahajan ?

Harsh Shrivastava - 5 years, 4 months ago

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@Harsh Shrivastava That's a good idea. But I have yet to learn whole integration , I have just finished the basics!

Nihar Mahajan - 5 years, 4 months ago

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@Nihar Mahajan That's why I suggested to have an age limit. But if parth and kirtarth participate then.... The contest will end then and there.

Aditya Kumar - 5 years, 4 months ago

Thank you :) Yes , we would hold it in April , till then I would have learnt some integration :P

Nihar Mahajan - 5 years, 4 months ago

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LOL would i be able to take part in junior contest(i am only 13).

Aareyan Manzoor - 5 years, 4 months ago

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@Aareyan Manzoor Why not :)

Harsh Shrivastava - 5 years, 4 months ago

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@Harsh Shrivastava Because i know advanced integration techniques(learned them right after the end of the integration contest).

Aareyan Manzoor - 5 years, 4 months ago

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@Aareyan Manzoor No worries :)

Harsh Shrivastava - 5 years, 4 months ago

It would've been better if you had given a simple proof of that property. It would be easy for beginners.

Aditya Kumar - 5 years, 4 months ago

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