Greatest Integer of Sum of Inverse Function Integrals

To find the greatest integer less than or equal to the expression $I = \int_1^2 \log_2(x^3 + 1) \, dx + \int_1^{\log_2 9} (2^x - 1)^{\frac{1}{3}} \, dx,$

we analyze the two integrals by using the property of inverse functions in definite integration.

Step 1: Identify the Inverse Function

Let $f(x) = \log_2(x^3 + 1)$.

To find the inverse function $f^{-1}(x)$, let $y = f(x)$: $y = \log_2(x^3 + 1)$ $2^y = x^3 + 1$ $x^3 = 2^y - 1$ $x = (2^y - 1)^{\frac{1}{3}}$

Thus, $f^{-1}(x) = (2^x - 1)^{\frac{1}{3}}$.

Step 2: Evaluate Limits of Integration

For the lower and upper limits of the first integral:

• At $x = 1$: $f(1) = \log_2(1^3 + 1) = \log_2 2 = 1$
• At $x = 2$: $f(2) = \log_2(2^3 + 1) = \log_2 9$

Thus, the given expression can be rewritten as: $I = \int_1^2 f(x) \, dx + \int_{f(1)}^{f(2)} f^{-1}(x) \, dx$

Step 3: Apply the Identity for Inverse Functions

Using the standard definite integration identity for a continuous strictly increasing function $f(x)$: $\int_a^b f(x) \, dx + \int_{f(a)}^{f(b)} f^{-1}(y) \, dy = b \cdot f(b) - a \cdot f(a)$

Substitute $a = 1$ and $b = 2$: $I = 2 \cdot f(2) - 1 \cdot f(1)$ $I = 2 \log_2 9 - 1 \cdot 1 = 2 \log_2 9 - 1$

Step 4: Estimate the Value of $I$

Simplifying $I$: $I = \log_2(9^2) - 1 = \log_2 81 - 1$

Since $2^6 = 64$ and $2^7 = 128$, we have: $64 < 81 < 128$ $\log_2 64 < \log_2 81 < \log_2 128$ $6 < \log_2 81 < 7$

Subtracting $1$ throughout: $6 - 1 < \log_2 81 - 1 < 7 - 1$ $5 < I < 6$

Conclusion

The greatest integer less than or equal to $I$ is: $\lfloor I \rfloor = 5$