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^(2))dx` then

Options

A. `I_(1) gt I_(2)` and `I_(4) gt I_(3)`
B. `I_(2)gt I_(1)` and `I_(3) gt I_(4)`
C. `I_(1) gt I_(2)` and `I_(3) gt I_(4)`
D. none of these

Answer

To solve the problem, we need to compare the integrals \( I_1, I_2, I_3, \) and \( I_4 \) as defined in the question. Let's go through the steps systematically.

### Step 1: Define the Integrals
We have the following integrals:
- \( I_1 = \int_{0}^{1} 2^{x^2} \, dx \)
- \( I_2 = \int_{0}^{1} 2^{x^3} \, dx \)
- \( I_3 = \int_{1}^{2} 2^{x^2} \, dx \)
- \( I_4 = \int_{1}^{2} 2^{x^3} \, dx \)

### Step 2: Compare \( I_1 \) and \( I_2 \)
To compare \( I_1 \) and \( I_2 \), we note that for \( x \) in the interval \([0, 1]\):
- \( x^2 \leq x^3 \) (since \( x \) is between 0 and 1)
This implies:
- \( 2^{x^2} \geq 2^{x^3} \)

Thus, we can write:
\[ I_1 = \int_{0}^{1} 2^{x^2} \, dx \geq \int_{0}^{1} 2^{x^3} \, dx = I_2 \]
So, we conclude:
\[ I_1 > I_2 \]

### Step 3: Compare \( I_3 \) and \( I_4 \)
Now, we compare \( I_3 \) and \( I_4 \). For \( x \) in the interval \([1, 2]\):
- \( x^2 \geq x^3 \) (since \( x \) is greater than 1)
This implies:
- \( 2^{x^2} \geq 2^{x^3} \)

Thus, we can write:
\[ I_3 = \int_{1}^{2} 2^{x^2} \, dx \geq \int_{1}^{2} 2^{x^3} \, dx = I_4 \]
So, we conclude:
\[ I_3 > I_4 \]

### Final Conclusion
From our comparisons, we have:
- \( I_1 > I_2 \)
- \( I_3 > I_4 \)

### Summary of Results
- \( I_1 > I_2 \)
- \( I_3 > I_4 \)