1. Evaluate the integral:

We start with: $I = \int_{0}^{1} \int_{0}^{y} e^{x-y} \, dx \, dy$

Step 1: Simplify the inner integral with respect to $x$:

The inner integral is: $\int_{0}^{y} e^{x-y} \, dx$ Since $e^{x-y} = e^{-y} \cdot e^{x}$, treat $e^{-y}$ as a constant: $\int_{0}^{y} e^{-y} e^{x} \, dx = e^{-y} \int_{0}^{y} e^{x} \, dx$

Now evaluate $\int_{0}^{y} e^{x} \, dx$: $\int_{0}^{y} e^{x} \, dx = \left[e^{x}\right]_{0}^{y} = e^{y} - e^{0} = e^{y} - 1$

Thus, the inner integral becomes: $\int_{0}^{y} e^{x-y} \, dx = e^{-y} (e^{y} - 1) = 1 - e^{-y}$

Step 2: Evaluate the outer integral with respect to $y$:

Substitute the result of the inner integral: $I = \int_{0}^{1} (1 - e^{-y}) \, dy$ Split the integral: $I = \int_{0}^{1} 1 \, dy - \int_{0}^{1} e^{-y} \, dy$

Evaluate each term:

1. $\int_{0}^{1} 1 \, dy = [y]_{0}^{1} = 1 - 0 = 1$
2. $\int_{0}^{1} e^{-y} \, dy = \left[-e^{-y}\right]_{0}^{1} = -(e^{-1} - e^{0}) = -(1/e - 1) = 1 - 1/e$

Combine the results: $I = 1 - (1 - 1/e) = \frac{1}{e}$

Final Result:

$I = \frac{1}{e}$

2. Reversing the order of integration:

The region of integration is defined by: $0 \leq x \leq y \quad \text{and} \quad 0 \leq y \leq 1$

Reversing the order of integration requires expressing the bounds in terms of $x$: $0 \leq y \leq 1 \quad \text{and} \quad x \leq y \leq 1$

The integral becomes: $I = \int_{0}^{1} \int_{x}^{1} e^{x-y} \, dy \, dx$

BONUS: Evaluate the line integral

The line integral is: $J = \int_{C} e^{2x - y^2} \, dx - y e^{2x - y^2} \, dy$ with the curve $C: \mathbf{r}(t) = (t^2, t), \, t \in [0, 1]$.

Step 1: Parameterize the integral

From $\mathbf{r}(t) = (t^2, t)$, we have: $x = t^2, \quad y = t, \quad dx = 2t \, dt, \quad dy = 1 \, dt$

Substitute into the integral: $J = \int_{0}^{1} \left[ e^{2t^2 - t^2}(2t) - t e^{2t^2 - t^2}(1) \right] \, dt$

Simplify: $J = \int_{0}^{1} \left[ 2t e^{t^2} - t e^{t^2} \right] \, dt = \int_{0}^{1} t e^{t^2} \, dt$

Step 2: Evaluate the integral

Use substitution: Let $u = t^2$, so $du = 2t \, dt$. Adjust the bounds:

• When $t = 0$, $u = 0$
• When $t = 1$, $u = 1$

The integral becomes: $\int_{0}^{1} t e^{t^2} \, dt = \frac{1}{2} \int_{0}^{1} e^{u} \, du$

Evaluate: $\frac{1}{2} \int_{0}^{1} e^{u} \, du = \frac{1}{2} \left[e^{u}\right]_{0}^{1} = \frac{1}{2} (e^{1} - e^{0}) = \frac{1}{2} (e - 1)$

Final Result:

$J = \frac{e - 1}{2}$

Do you have any questions or need further clarification?

Related Questions:

1. How can the region of integration be visualized geometrically?
2. What are the benefits of reversing the order of integration?
3. How does parameterization simplify line integrals in vector calculus?
4. Can $e^{x-y}$ integrals be extended to higher dimensions?
5. What alternative methods exist for evaluating the line integral $J$?

Tip:

Always sketch the region of integration for double integrals; it helps in correctly reversing the integration bounds!