The greatest integer less than or equal to
[tex] \rm \int_{1}^2 log_{2}( {x}^{3} + 1 ) dx + \int_{ 1 }^{log_{2}(9)} ( {2}^x - 1) {}^{ \frac{1}{3} } \: dx \: is \\ [/tex]

Respuesta :

We use a property from an earlier question of yours [27174924], that

[tex]\displaystyle \int_a^b f(x) \, dx + \int_{f(a)}^{f(b)} f^{-1}(x) \, dx = b\,f(b) - a\, f(a)[/tex]

Note that

[tex]f(x) = \log_2(x^3+1) \implies f^{-1}(x) = \left(2^x - 1\right)^{1/3}[/tex]

[tex]a = 1 \implies f(a) = \log_2(1^3 + 1) = \log_2(2) = 1[/tex]

[tex]b = 2 \implies f(b) = \log_2(2^3 + 1) = \log_2(9)[/tex]

so that the exact value of the integral is

[tex]\displaystyle \int_1^2 \log_2(x^3 + 1) \, dx + \int_1^{\log_2(9)} \left(2^x - 1\right)^{1/3} = 2\log_2(9) - 1[/tex]

Now, observe that

[tex]8 = 2^3 < 9 < 2^{7/2} = 8\sqrt2 \approx11.3 \\\\ \implies 3 < \log_2(9) < \dfrac72 \\\\ \implies 6 < 2\log_2(9) < 7 \\\\ \implies 5 < 2\log_2(9) - 1 < 6[/tex]

so that

[tex]\left\lfloor 2\log_2(9) - 1 \right\rfloor = \boxed{5}[/tex]