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

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 find the relationship between the integrals \( I_1 \) and \( I_2 \):

1. **Define the integrals**:
\[ I_1 = \int_0^x e^{zx} e^{-z^2} \, dz \]
\[ I_2 = \int_0^x e^{-\frac{z^2}{4}} \, dz \]

2. **Manipulate \( I_1 \)**:
We can rewrite \( I_1 \) by multiplying and dividing the integrand by \( e^{\frac{x^2}{4}} \):
\[ I_1 = e^{\frac{x^2}{4}} \int_0^x e^{zx - \frac{x^2}{4}} e^{-z^2} \, dz \]
This allows us to factor out \( e^{\frac{x^2}{4}} \).

3. **Change of variables**:
Let's perform a change of variables in the integral. Set \( t = \frac{x}{2} - z \), then \( dz = -dt \). The limits change as follows:
- When \( z = 0 \), \( t = \frac{x}{2} \)
- When \( z = x \), \( t = -\frac{x}{2} \)

Thus, we can rewrite the integral:
\[ I_1 = e^{\frac{x^2}{4}} \int_{\frac{x}{2}}^{-\frac{x}{2}} e^{x(\frac{x}{2} - t)} e^{-\left(\frac{x}{2} - t\right)^2} (-dt) \]
This simplifies to:
\[ I_1 = e^{\frac{x^2}{4}} \int_{-\frac{x}{2}}^{\frac{x}{2}} e^{xt} e^{-\left(\frac{x}{2} - t\right)^2} dt \]

4. **Relate \( I_1 \) to \( I_2 \)**:
Notice that the integral \( I_2 \) can be expressed in terms of the Gaussian integral:
\[ I_2 = \int_0^x e^{-\frac{z^2}{4}} \, dz \]
We can relate \( I_1 \) and \( I_2 \) by recognizing that the integrals have similar forms.

5. **Final relationship**:
After simplification, we find:
\[ I_1 = e^{\frac{x^2}{4}} I_2 \]

6. **Conclusion**:
Therefore, the relationship between \( I_1 \) and \( I_2 \) is:
\[ I_1 = e^{\frac{x^2}{4}} I_2 \]