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Solve $\int\limits_1^3 {\frac{{\left[ {{x^2}} \right]}}{{{{\left[ {{x^2} - 8x + 16} \right]}} + \left[ {{x^2}} \right]}}dx} = \_\_\_\_\_$ where [.] represent greatest integer function.

My approch is as follow

$\int\limits_1^3 {\frac{{\left[ {{x^2}} \right]}}{{\left[ {{{\left( {x - 4} \right)}^2}} \right] + \left[ {{x^2}} \right]}}dx} $

$y=(x-4)^2$ represent a quadratic equation Refer to the image below

enter image description here

We see that for some x value y will have values of 9,8,7,....,1 SO how we will proceed this

I presume it will be a telescopic function

Albus Dumbledore
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Samar Imam Zaidi
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  • Just break the integral at each point where one of the greatest integer functions changes value. The integrand is constant between those points. Where does $(x-4)^2$ drop below $8$? What is the value of the integrand below that? Above that? – Ross Millikan Sep 13 '21 at 03:57
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    use kings property replace $x\to 4-x$ and add the equivalent integrals – Albus Dumbledore Sep 13 '21 at 04:02

1 Answers1

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$$I=\int\limits_1^3 {\frac{{\left[ {{x^2}} \right]}}{{{{\left[ {(x-4)^2} \right]}} + \left[ {{x^2}} \right]}}dx}!$$ Use $$\int_{a}^{b} f(x) dx=\int_{a}^{b} f(a+b-x) dx$$, then $$I=\int\limits_1^3 {\frac{{\left[ {{(x-4)^2}} \right]}}{{{{\left[ {x^2} \right]}} + \left[ {{(x-4)^2}} \right]}}dx}$$ Adding the two we get $$2I=\int_{1}^{3} dx \implies I=1.$$

Z Ahmed
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