Question

The greatest integer less than or equal to\(\int\limits_1^2 \log _2\left(x^3+1\right) d x+\int\limits_1^{\log _2 9}\left(2^x-1\right)^{\frac{1}{3}} dx\) is _____

Answer 1

Correct Answer: 5

Mathematical Solution Using Logarithms 

The function \( f(x) \) is defined as: \[ f(x) = \log_2 \left( x^3 + 1 \right) = y \]

We can rearrange this equation to express \( x \) in terms of \( y \): \[ x^3 + 1 = 2^y \Rightarrow x = (2^y - 1)^{1/3} = f^{-1}(y) \]

The inverse of \( f(x) \) is given by: \[ f^{-1}(x) = (2^x - 1)^{1/3} \]

Now, integrating: \[ \int_1^2 \log_2 \left( x^3 + 1 \right) dx + \int_1^{\log_2 9} \left( (2^x - 1)^{1/3} \right) dx \]

This becomes: \[ \int_1^2 f(x) dx + \int_1^{\log_2 9} f^{-1}(x) dx = 2 \log_2 9 - 1 \]

Now simplifying: \[ 8 < 9 < 2^{7/2} \Rightarrow 3 < \log_2 9 < \frac{7}{2} \]

Therefore: \[ 5 < 2 \log_2 9 - 1 < 6 \]

Finally: \[ [2 \log_2 9 - 1] = 5 \]