Step-by-step Solution:

First, we will take the interval \( I_1 \) and \( I_2 \) because they have the same interval 0 to 1. Square of any value from the interval 0 to 1 is greater than the cube of that value. $$ \Rightarrow x^2 > x^3 \quad \text{in the interval } [0, 1] $$ Also, we can see that: $$ \int_0^1 x^2 > \int_0^1 x^3 \quad \text{in the interval } [0, 1] $$ We can also see that: $$ 2^{x^2} > 2^{x^3} \quad \text{in the interval } [0, 1] $$ Therefore, the value of \( \int_0^1 2^{x^2} , dx\) is greater than the value of \( \int_0^1 2^{x^3} \, dx \). $$ \Rightarrow \int_0^1 2^{x^2} \, dx > \int_0^1 2^{x^3} \, dx $$ Hence, \( I_1 > I_2\) --- Now, we can see that the cube of any value in the interval [1, 2] is greater than the square of that value. The value of \( x^3 \) in the interval [1, 2] is larger than the value of \( x^2 \). $$ \text{Hence, } x^3 > x^2 \quad \text{in the interval } [1, 2] $$ The integration of \( x^3 \) is greater than the integration of \( x^2 \) in the interval [1, 2]. $$ \Rightarrow \int_1^2 x^3 > \int_1^2 x^2 $$ Therefore, the value of \( \int_1^2 2^{x^3} \, dx \) is greater than the value of \( \int_1^2 2^{x^2} \, dx\) $$ \Rightarrow \int_1^2 2^{x^3} \, dx > \int_1^2 2^{x^2} \, dx $$ Hence, \( I_4 > I_3 \) Therefore, \( I_3 \ne I_4 \) ---