Integration of the greatest integer function 1

Calculus Level 5

I = 0 1 ( 1 ) 1 x d x \large I=\int\limits_0^1 \left(-1\right)^{^{\left\lfloor\frac{1}{x}\right\rfloor}} dx

Submit the value of I |I| to 3 decimal places.


The answer is 0.386.

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1 solution

Rimson Junio
Aug 21, 2015

The given integral is equivalent to: I = 1 2 1 ( 1 ) 1 d x + 1 3 1 2 ( 1 ) 2 d x + 1 4 1 3 ( 1 ) 3 d x + 1 5 1 4 ( 1 ) 4 d x + . . . I=\int\limits_\frac{1}{2}^1\left(-1\right)^1dx+\int\limits_\frac{1}{3}^\frac{1}{2}\left(-1\right)^2dx+\int\limits_\frac{1}{4}^\frac{1}{3}\left(-1\right)^3dx+\int\limits_\frac{1}{5}^\frac{1}{4}\left(-1\right)^4dx+... This simplifies to: I = ( 1 + 1 2 ) + ( 1 2 1 3 ) + ( 1 3 + 1 4 ) + ( 1 4 1 5 ) + . . . I=(-1+\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+ (-\frac{1}{3}+\frac{1}{4})+(\frac{1}{4}-\frac{1}{5})+... I = 1 + 2 ( 1 2 1 3 + 1 4 1 5 + . . . ) I=-1+2(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...) I = 1 + 2 ( 1 l n 2 ) = 0.3863... I=-1+2(1-ln2)=-0.3863... Finally, I = 0.3863... |I|=0.3863...

Isn't : -1+1-1+1-1+1-1+1-1....= -1/2 ?

Rikin Shah - 3 years, 5 months ago

I have approached the question in the same way you did.

Taisanul Haque - 2 years, 7 months ago

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