Animated Solution for Mathematics - Definite Integration: If $I_1={\int }_0^1{2}^{x^2}dx,I_2={\int }_0^1{2}^{x^3}dx,I_3={\int }_1^2{2}^{x^2}dx$ and $I_4={\int }_1^2{2}^{x^3}dx$ then

Welcome, future engineers! Today, we are diving into a problem that, at first glance, might make your heart race. You see four integrals:

Your instinct might be to reach for your pen and start calculating. But stop! In the world of JEE Advanced, the most powerful tool in your arsenal is not just calculation—it is intuition and visualization.

We are dealing with two functions: $f(x)=2^{x^2}$ and $g(x)=2^{x^3}$. These are exponential functions with a base of $2$.

Because the base $2$ is greater than $1$, the function $h(t)=2^t$ is strictly increasing. This is a crucial realization. It means that if we can show that one exponent is larger than another, the entire function value follows suit.

If $t_1>t_2$, then $2^{t_1}>2^{t_2}$. We don't need to integrate; we just need to compare the exponents $x^2$ and $x^3$ within the given intervals.

Let us look at the interval $(0,1)$. Imagine you are standing on the number line between $0$ and $1$. Pick a number, say $0.5$.

If you square it, you get $0.25$. If you cube it, you get $0.125$. Notice that $0.25>0.125$.

This is the secret of the fractional domain: higher powers make numbers smaller. Thus, for any $x\in (0,1)$, we have $x^2>x^3$.

Since our base $2>1$, this implies $2^{x^2}>2^{x^3}$. Graphically, the curve $y=2^{x^2}$ sits comfortably above $y=2^{x^3}$ in this region. Because the area under the higher curve is larger, we immediately conclude that $I_1>I_2$.

Now, let us cross the threshold at $x=1$. Here, the behavior of powers flips entirely. Take a number like $1.5$.

Squaring it gives $2.25$, but cubing it gives $3.375$. Now, the higher power is the winner! For any $x\in (1,2)$, we have $x^3>x^2$.

Applying our increasing function property again, we get $2^{x^3}>2^{x^2}$. The curve $y=2^{x^3}$ has now soared above the curve $y=2^{x^2}$.

Consequently, the area under the curve $y=2^{x^3}$ from $1$ to $2$ must be larger than the area under the curve $y=2^{x^2}$. This gives us our second conclusion: $I_4>I_3$.

We have navigated the intervals and compared the functions without ever needing to perform a complex integration. We found that $I_1>I_2$ and $I_4>I_3$.

This problem is a beautiful reminder that in physics and mathematics, the most elegant solution is often the one that relies on understanding the nature of the functions rather than brute-force calculation. Keep this perspective, and you will conquer any problem the JEE throws your way!