We need to find the mass $m$ and center of mass $(\bar{x}, \bar{y})$ of the lamina bounded by the curves $y = e^x$, $y = 0$, $x = 0$, $x = 1$, with density function $\delta(x, y) = 2 - x + y$.

1. Problem Description

We need to find the mass $m$ and center of mass $(\bar{x}, \bar{y})$ of the lamina bounded by the curves $y = e^x$, $y = 0$, $x = 0$, $x = 1$, with density function $\delta(x, y) = 2 - x + y$.

2. Solution Steps

First, we calculate the mass $m$ of the lamina:

$$m = \int\int_R \delta(x, y) \, dA = \int_0^1 \int_0^{e^x} (2 - x + y) \, dy \, dx$$

We first evaluate the inner integral with respect to $y$:

$$\int_0^{e^x} (2 - x + y) \, dy = \left[(2 - x)y + \frac{1}{2}y^2\right]_0^{e^x} = (2 - x)e^x + \frac{1}{2}e^{2x}$$

Now, we evaluate the outer integral with respect to $x$:

$$m = \int_0^1 \left[(2 - x)e^x + \frac{1}{2}e^{2x}\right] \, dx = \int_0^1 (2 - x)e^x \, dx + \frac{1}{2} \int_0^1 e^{2x} \, dx$$

We evaluate $\int_0^1 (2 - x)e^x \, dx$ using integration by parts: Let $u = 2 - x$, $dv = e^x \, dx$, so $du = -dx$, $v = e^x$. Then,

$$\int_0^1 (2 - x)e^x \, dx = \left[(2 - x)e^x\right]_0^1 - \int_0^1 e^x (-dx) = (2 - 1)e^1 - (2 - 0)e^0 + \int_0^1 e^x \, dx = e - 2 + [e^x]_0^1 = e - 2 + e - 1 = 2e - 3$$

We evaluate $\frac{1}{2} \int_0^1 e^{2x} \, dx$:

$$\frac{1}{2} \int_0^1 e^{2x} \, dx = \frac{1}{2} \left[\frac{1}{2}e^{2x}\right]_0^1 = \frac{1}{4}(e^2 - 1)$$

Therefore,

$$m = 2e - 3 + \frac{1}{4}(e^2 - 1) = 2e - 3 + \frac{1}{4}e^2 - \frac{1}{4} = \frac{1}{4}e^2 + 2e - \frac{13}{4}$$

Next, we calculate $M_y$ and $M_x$:

$$M_y = \int\int_R x\delta(x, y) \, dA = \int_0^1 \int_0^{e^x} x(2 - x + y) \, dy \, dx$$

$$\int_0^{e^x} x(2 - x + y) \, dy = x\left[(2 - x)y + \frac{1}{2}y^2\right]_0^{e^x} = x(2 - x)e^x + \frac{1}{2}xe^{2x} = (2x - x^2)e^x + \frac{1}{2}xe^{2x}$$

$$M_y = \int_0^1 \left[(2x - x^2)e^x + \frac{1}{2}xe^{2x}\right] \, dx = \int_0^1 (2x - x^2)e^x \, dx + \frac{1}{2} \int_0^1 xe^{2x} \, dx$$

We evaluate $\int_0^1 (2x - x^2)e^x \, dx$ using integration by parts twice:

$\int x^2 e^x dx = x^2 e^x - \int 2x e^x dx = x^2 e^x - 2(xe^x - e^x)$

$\int 2x e^x dx = 2 \int xe^x dx = 2(xe^x - e^x)$

So $\int (2x-x^2)e^x dx = 2(xe^x - e^x) - (x^2 e^x - 2xe^x + 2e^x) = (4x-x^2)e^x-4e^x$

$\int_0^1 (2x-x^2)e^x dx = (4-1)e-4e - (0-4) = -e + 4$

We evaluate $\frac{1}{2}\int_0^1 x e^{2x}dx$ using integration by parts:

$u=x$, $dv = e^{2x} dx$, so $du=dx$ $v = \frac{1}{2} e^{2x}$

Then, $\frac{1}{2} \int_0^1 xe^{2x} dx = \frac{1}{2}[x (\frac{1}{2}e^{2x})|_0^1 - \int_0^1 \frac{1}{2}e^{2x} dx] = \frac{1}{2}[ \frac{1}{2}e^2 - \frac{1}{2} (\frac{1}{2}e^{2x})|_0^1] = \frac{1}{4}e^2 - \frac{1}{4}(\frac{1}{2}e^2 - \frac{1}{2}) = \frac{1}{4}e^2 - \frac{1}{8}e^2 + \frac{1}{8} = \frac{1}{8}e^2 + \frac{1}{8}$

$M_y = -e+4 + \frac{1}{8}e^2 + \frac{1}{8} = \frac{1}{8}e^2 - e + \frac{33}{8}$

$\bar{x} = \frac{M_y}{m} = \frac{\frac{1}{8}e^2 - e + \frac{33}{8}}{\frac{1}{4}e^2 + 2e - \frac{13}{4}} = \frac{e^2 - 8e + 33}{2e^2 + 16e - 26}$

$$M_x = \int\int_R y\delta(x, y) \, dA = \int_0^1 \int_0^{e^x} y(2 - x + y) \, dy \, dx$$

$$\int_0^{e^x} y(2 - x + y) \, dy = \int_0^{e^x} (2y - xy + y^2) \, dy = \left[y^2 - \frac{1}{2}xy^2 + \frac{1}{3}y^3\right]_0^{e^x} = e^{2x} - \frac{1}{2}xe^{2x} + \frac{1}{3}e^{3x}$$

$$M_x = \int_0^1 \left[e^{2x} - \frac{1}{2}xe^{2x} + \frac{1}{3}e^{3x}\right] \, dx = \int_0^1 e^{2x} dx - \frac{1}{2}\int_0^1 xe^{2x} dx + \frac{1}{3}\int_0^1 e^{3x} dx$$

We know that $\int_0^1 e^{2x} dx = \frac{1}{2}e^2 - \frac{1}{2}$ and $\int_0^1 xe^{2x} dx = \frac{1}{4}e^2 + \frac{1}{4}$, and $\int_0^1 e^{3x}dx = \frac{1}{3} e^3 - \frac{1}{3}$.

So $M_x = (\frac{1}{2}e^2 - \frac{1}{2}) - \frac{1}{2} (\frac{1}{4}e^2 + \frac{1}{4}) + \frac{1}{3}(\frac{1}{3}e^3 - \frac{1}{3}) = \frac{1}{2}e^2 - \frac{1}{2} - \frac{1}{8}e^2 - \frac{1}{8} + \frac{1}{9}e^3 - \frac{1}{9} = \frac{1}{9}e^3 + \frac{3}{8}e^2 - \frac{49}{72}$

$\bar{y} = \frac{M_x}{m} = \frac{\frac{1}{9}e^3 + \frac{3}{8}e^2 - \frac{49}{72}}{\frac{1}{4}e^2 + 2e - \frac{13}{4}} = \frac{8e^3 + 27e^2 - 49}{18e^2 + 144e - 117}$

3. Final Answer

$m = \frac{1}{4}e^2 + 2e - \frac{13}{4}$

$\bar{x} = \frac{e^2 - 8e + 33}{2e^2 + 16e - 26}$

$\bar{y} = \frac{8e^3 + 27e^2 - 49}{18e^2 + 144e - 117}$