If `I_(1)=int_(0)^(x) e^("zx "e^(-z^(2)))dz` and `I_(2)=int_(0)^(x) e^(-z^(2)//4)dz`, then

Options

A. `I_(1)=e^(x)I_(2)`
B. `I_(1)=e^(x^(2))I_(2)`
C. `I_(1)=e^(x^(2)//2)I_(2)`
D. none of these

Answer

To solve the problem, we need to evaluate the integrals \( I_1 \) and \( I_2 \) given by:

\[ I_1 = \int_0^x e^{zx} e^{-z^2} \, dz \]

\[ I_2 = \int_0^x e^{-\frac{z^2}{4}} \, dz \]

### Step 1: Rewrite \( I_1 \)

We can rewrite \( I_1 \) by factoring out \( e^{\frac{x^2}{4}} \):

\[ I_1 = e^{\frac{x^2}{4}} \int_0^x e^{-\left(z - \frac{x}{2}\right)^2} \, dz \]

### Step 2: Change of Variables in \( I_1 \)

Let’s perform a change of variables in the integral. Set:

\[ t = z - \frac{x}{2} \implies z = t + \frac{x}{2} \]

Then, the limits change as follows:
- When \( z = 0 \), \( t = -\frac{x}{2} \)
- When \( z = x \), \( t = \frac{x}{2} \)

The differential \( dz \) becomes \( dt \):

\[ I_1 = e^{\frac{x^2}{4}} \int_{-\frac{x}{2}}^{\frac{x}{2}} e^{-t^2} \, dt \]

### Step 3: Relate \( I_1 \) to \( I_2 \)

Notice that \( I_2 \) can be expressed in terms of the same limits:

\[ I_2 = \int_0^x e^{-\frac{z^2}{4}} \, dz \]

To relate \( I_1 \) and \( I_2 \), we can use the properties of integrals. The integral \( I_2 \) can be rewritten as:

\[ I_2 = \int_0^x e^{-\frac{z^2}{4}} \, dz = 2 \int_0^{\frac{x}{2}} e^{-u^2} \, du \quad \text{(where } u = \frac{z}{2} \text{)} \]

### Step 4: Final Expression

Thus, we can express \( I_1 \) in terms of \( I_2 \):

\[ I_1 = e^{\frac{x^2}{4}} \cdot I_2 \]

### Conclusion

The relationship between \( I_1 \) and \( I_2 \) is:

\[ I_1 = e^{\frac{x^2}{4}} I_2 \]