1. Key Concepts and Formulas

• Comparison Property of Definite Integrals: If $f(x)$ and $g(x)$ are integrable on $[a, b]$ and $f(x) \ge g(x)$ for all $x \in [a, b]$, then $\int_a^b f(x) \,dx \ge \int_a^b g(x) \,dx$. If $f(x) > g(x)$ on a subinterval of non-zero length, then $\int_a^b f(x) \,dx > \int_a^b g(x) \,dx$.
• Behavior of Exponential Functions: The function $y = 2^u$ is an increasing function of $u$.
• Properties of Powers: For $x > 0$, $x^2 \ge x^3$ if and only if $x \le 1$. For $x > 0$, $x^3 \ge x^2$ if and only if $x \ge 1$.

2. Step-by-Step Solution

Step 1: Analyze the integrands and intervals for $I_1$ and $I_2$. We are given $I_1 = \int_0^1 2^{x^2} \,dx$ and $I_2 = \int_0^1 2^{x^3} \,dx$. Both integrals are over the interval $[0, 1]$. We need to compare the exponents $x^2$ and $x^3$ for $x \in [0, 1]$.

Reasoning: To compare the integrals, we will compare their integrands. Since the base $2$ is greater than $1$, the function $2^u$ is an increasing function. Therefore, if the exponent of one integrand is greater than or equal to the exponent of the other integrand, the value of the integrand will also be greater than or equal to.

Step 2: Compare the exponents $x^2$ and $x^3$ on the interval $[0, 1]$. For $x \in [0, 1]$, we can analyze the relationship between $x^2$ and $x^3$. Consider the difference $x^2 - x^3 = x^2(1 - x)$. For $x \in [0, 1]$:

• $x^2 \ge 0$.
• $1 - x \ge 0$. Therefore, $x^2(1 - x) \ge 0$ for all $x \in [0, 1]$. This implies $x^2 \ge x^3$ for all $x \in [0, 1]$. Furthermore, for $x \in (0, 1)$, $x^2 > x^3$. For example, if $x = 0.5$, then $x^2 = 0.25$ and $x^3 = 0.125$, so $x^2 > x^3$.

Reasoning: Understanding the behavior of powers of $x$ in the interval $[0, 1]$ is crucial. For $x$ between $0$ and $1$, higher powers of $x$ result in smaller values.

Step 3: Apply the Comparison Property to compare $I_1$ and $I_2$. Since $x^2 \ge x^3$ for all $x \in [0, 1]$, and $x^2 > x^3$ for $x \in (0, 1)$ (an interval of non-zero length), we can compare the integrands $2^{x^2}$ and $2^{x^3}$. Because $2^u$ is an increasing function, if $a \ge b$, then $2^a \ge 2^b$. Thus, $2^{x^2} \ge 2^{x^3}$ for all $x \in [0, 1]$. Since $2^{x^2} > 2^{x^3}$ on the interval $(0, 1)$, by the Comparison Property of Definite Integrals: $\int_0^1 2^{x^2} \,dx > \int_0^1 2^{x^3} \,dx$ Therefore, $I_1 > I_2$.

Reasoning: The Comparison Property directly allows us to relate the integral values based on the relative magnitudes of their integrands over the given interval.

Step 4: Analyze the integrands and intervals for $I_3$ and $I_4$. We are given $I_3 = \int_1^2 2^{x^2} \,dx$ and $I_4 = \int_1^2 2^{x^3} \,dx$. Both integrals are over the interval $[1, 2]$. We need to compare the exponents $x^2$ and $x^3$ for $x \in [1, 2]$.

Reasoning: Similar to the previous comparison, we will compare the integrands by comparing their exponents.

Step 5: Compare the exponents $x^2$ and $x^3$ on the interval $[1, 2]$. For $x \in [1, 2]$, we can analyze the relationship between $x^2$ and $x^3$. Consider the difference $x^3 - x^2 = x^2(x - 1)$. For $x \in [1, 2]$:

• $x^2 > 0$.
• $x - 1 \ge 0$. Therefore, $x^2(x - 1) \ge 0$ for all $x \in [1, 2]$. This implies $x^3 \ge x^2$ for all $x \in [1, 2]$. Furthermore, for $x \in (1, 2]$, $x^3 > x^2$. For example, if $x = 1.5$, then $x^2 = 2.25$ and $x^3 = 3.375$, so $x^3 > x^2$.

Reasoning: For $x > 1$, higher powers of $x$ result in larger values.

Step 6: Apply the Comparison Property to compare $I_3$ and $I_4$. Since $x^3 \ge x^2$ for all $x \in [1, 2]$, and $x^3 > x^2$ for $x \in (1, 2]$ (an interval of non-zero length), we can compare the integrands $2^{x^3}$ and $2^{x^2}$. Because $2^u$ is an increasing function, if $a \ge b$, then $2^a \ge 2^b$. Thus, $2^{x^3} \ge 2^{x^2}$ for all $x \in [1, 2]$. Since $2^{x^3} > 2^{x^2}$ on the interval $(1, 2]$, by the Comparison Property of Definite Integrals: $\int_1^2 2^{x^3} \,dx > \int_1^2 2^{x^2} \,dx$ Therefore, $I_4 > I_3$.

Reasoning: The Comparison Property is applied again to determine the relationship between $I_3$ and $I_4$.

Step 7: Evaluate the given options based on our findings. We found that $I_1 > I_2$. Let's check the options:

• (A) $I_2 > I_1$: This is false.
• (B) $I_1 > I_2$: This is true.
• (C) $I_3 = I_4$: This is false, as we found $I_4 > I_3$.
• (D) $I_3 > I_4$: This is false, as we found $I_4 > I_3$.

Reasoning: We systematically check each option against the conclusions derived from our analysis.

3. Common Mistakes & Tips

• Confusing Interval Behavior: Be extremely careful about the interval of integration. The relationship between $x^2$ and $x^3$ (and thus between $2^{x^2}$ and $2^{x^3}$) is reversed for $x \in [0, 1]$ compared to $x \in [1, \infty)$.
• Incorrectly Applying Comparison Property: Ensure that you are comparing the correct functions and that the inequality of the integrands holds over the entire interval of integration.
• Assuming Integrability: The functions $2^{x^2}$ and $2^{x^3}$ are continuous, and thus integrable, on closed intervals. This is a prerequisite for using the Comparison Property.

4. Summary

To compare the definite integrals $I_1$ and $I_2$, we analyzed the exponents $x^2$ and $x^3$ on the interval $[0, 1]$. For $x \in [0, 1]$, we found that $x^2 \ge x^3$. Since the base $2$ is greater than $1$, the function $2^u$ is increasing, which implies $2^{x^2} \ge 2^{x^3}$ on $[0, 1]$. As this inequality is strict on $(0, 1)$, by the Comparison Property of Definite Integrals, $I_1 > I_2$. For $I_3$ and $I_4$, over the interval $[1, 2]$, we found $x^3 \ge x^2$, leading to $2^{x^3} \ge 2^{x^2}$, and thus $I_4 > I_3$. Comparing these results with the given options, option (B) $I_1 > I_2$ is the correct statement.

5. Final Answer

The final answer is $\boxed{A}$.