AF(x,y) = ∫ex^2[x+y]dyBF(x,y) = ∫ ex^2[x - y] dyCF(x,y) = ∫ ex^2)[x2 + y2] dyDF(x,y) = ∫ ex^2[x2 – y2] dy

To analyze the given integrals, we will investigate them based on the structure provided. Since no specific question or problem is attached, I'll present the integrals in their respective forms and outline a general approach for evaluating them.

1. Break Down the Problem

   • Identify each integral clearly:
     • $AF(x, y) = \int e^{x^2} (x + y) \, dy$
     • $BF(x, y) = \int e^{x^2} (x - y) \, dy$
     • $CF(x, y) = \int e^{x^2} (x^2 + y^2) \, dy$
     • $DF(x, y) = \int e^{x^2} (x^2 - y^2) \, dy$

2. Relevant Concepts

   • Each of the integrals involves the exponential function $e^{x^2}$ multiplied by a polynomial in $y$. Basic integration principles and knowledge of exponential functions will apply.

3. Analysis and Detail

   • The integrals can be evaluated in terms of polynomial expressions:

     • For $AF(x, y)$: $AF(x, y) = e^{x^2} \int (x + y) \, dy = e^{x^2} \left( xy + \frac{y^2}{2} + C_1 \right)$

     • For $BF(x, y)$: $BF(x, y) = e^{x^2} \int (x - y) \, dy = e^{x^2} \left( xy - \frac{y^2}{2} + C_2 \right)$

     • For $CF(x, y)$: $CF(x, y) = e^{x^2} \int (x^2 + y^2) \, dy = e^{x^2} \left( x^2y + \frac{y^3}{3} + C_3 \right)$

     • For $DF(x, y)$: $DF(x, y) = e^{x^2} \int (x^2 - y^2) \, dy = e^{x^2} \left( x^2y - \frac{y^3}{3} + C_4 \right)$

4. Verify and Summarize

   • Each integral was simplified and evaluated based on its polynomial components over $y$, with exponential terms retained outside due to their independence from $y$.

Final Answer

1. $AF(x, y) = e^{x^2} \left( xy + \frac{y^2}{2} + C_1 \right)$
2. $BF(x, y) = e^{x^2} \left( xy - \frac{y^2}{2} + C_2 \right)$
3. $CF(x, y) = e^{x^2} \left( x^2y + \frac{y^3}{3} + C_3 \right)$
4. $DF(x, y) = e^{x^2} \left( x^2y - \frac{y^3}{3} + C_4 \right)$

Here, $C_1, C_2, C_3,$ and $C_4$ are constants of integration.