Is this Fundamental Theorem of Calculus?

Calculus Level 2

Let f ( x ) = ∫ 0 x t ( 2 t + 1 ) 2 t − 1 d t \displaystyle f(x) = \int_0^x \frac{t(2^t + 1)}{2^t - 1} \, dt .

What is the value of f ( 1 ) + f ( − 1 ) f(1) + f(-1) ?


The answer is 0.

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Arpita Karkera by Arpita Karkera , 24, India

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1 solution

Arpita Karkera
May 14, 2015

( 2 t + 1 ) t 2 t − 1 { \frac { ({ 2 }^{ t }+1)t }{ { 2 }^{ t }-1 } } is an even function.

Therefore, f(x) is and odd function and f(x) + f(-x)=0

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Creative question and answer! Bonus question: what is the easiest method to calculate f ( 1 ) − f ( − 1 ) f(1) - f(-1) ?

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