Animated Solution for Mathematics - Definite Integration: The greatest integer less than or equal to ${\int }_1^2{\log }_2(x^3+1)dx+{\int }_1^{{\log }_{2}9}(2^x-1{)}^{1/3}dx$ is ______.

Welcome, student. Today, we are going to dismantle a problem that, at first glance, might make your heart skip a beat.

You see two integrals, ${\int }_1^2{\log }_2(x^3+1)dx$ and ${\int }_1^{{\log }_{2}9}(2^x-1{)}^{\frac{1}{3}}dx$. Your instinct might be to reach for integration by parts or some complex substitution.

But pause. In the world of JEE Advanced, when you see two integrals that look like they do not belong together, there is almost always a hidden symmetry waiting to be discovered. Let us embark on this journey to find it.

The Detective Work

Let us define our primary function as $f(x)={\log }_2(x^3+1)$. If we try to integrate this directly, we will find ourselves in a labyrinth of non-elementary functions.

Instead, let us play the role of a detective. What if the second integrand is related to the first? Let us find the inverse of $f(x)$.

We set $y={\log }_2(x^3+1)$. Converting this to exponential form, we get:

Rearranging for $x$, we find $x^3=2^y-1$, which means $x=(2^y-1{)}^{\frac{1}{3}}$. If we swap $x$ and $y$, we get $f^{-1}(x)=(2^x-1{)}^{\frac{1}{3}}$.

Look at that! It is an exact match for the second integrand. This is not a coincidence; it is the signature of a well-crafted problem.

The Geometric Revelation

Now, let us verify the limits. For the first integral, the limits are $a=1$ and $b=2$.

Let us calculate the function values at these boundaries:

The second integral is defined from $1$ to ${\log }_{2}9$, which are exactly $f(1)$ and $f(2)$. We are looking at the sum of the area under $f(x)$ from $a$ to $b$ and the area under $f^{-1}(x)$ from $f(a)$ to $f(b)$.

Geometrically, this is the classic 'L-shape' area. The area under $f(x)$ is the region bounded by the curve and the $x$-axis. The area under $f^{-1}(x)$ is the region bounded by the curve and the $y$-axis.

When you add these two areas together, you are essentially filling the rectangle of dimensions $b\times f(b)$ and subtracting the smaller rectangle of dimensions $a\times f(a)$. The formula is elegant:

The Calculation

With this powerful tool in our arsenal, the integration vanishes. We simply plug in our values:

Substituting the values we found earlier:

Using the power rule of logarithms, $2{\log }_2(9)={\log }_2(9^2)={\log }_2(81)$. Thus, our integral simplifies to:

We have reduced a terrifying calculus problem to a simple logarithmic expression. This is the beauty of mathematical intuition.

The Final Estimation

Finally, we need the greatest integer less than or equal to $I$. We know that $2^6=64$ and $2^7=128$.

Since $64<81<128$, it follows that:

This implies $6<{\log }_2(81)<7$. Subtracting $1$ from all sides, we get:

Therefore, $5<I<6$. The greatest integer less than or equal to $I$ is clearly 5.

You have successfully navigated the trap, visualized the geometry, and arrived at the solution with elegance. Keep this geometric perspective in your toolkit; it will serve you well in the exam hall.