1. **Problem Statement:** Find the area of the shaded region bounded by the curves $y = e^x$, $y = xe^x$, and the vertical line $x = 2$. 2. **Understanding the curves:** - $y = e^x$ is an exponential function starting at $(0,1)$ and increasing. - $y = xe^x$ starts at $(0,0)$ and grows faster than $e^x$ for $x > 1$. 3. **Determine the interval:** The region is bounded between $x=0$ and $x=2$. 4. **Find the points of intersection:** At $x=0$, $e^0=1$ and $0 \cdot e^0=0$, so $y=e^x$ is above $y=xe^x$. 5. **Set up the integral for the area:** The area between two curves $y=f(x)$ and $y=g(x)$ from $a$ to $b$ is $$\text{Area} = \int_a^b |f(x) - g(x)| \, dx$$ Here, $f(x) = e^x$ and $g(x) = xe^x$, and since $e^x > xe^x$ on $[0,2]$, $$\text{Area} = \int_0^2 (e^x - xe^x) \, dx$$ 6. **Simplify the integrand:** $$e^x - xe^x = e^x(1 - x)$$ 7. **Compute the integral:** $$\int_0^2 e^x(1 - x) \, dx = \int_0^2 e^x \, dx - \int_0^2 x e^x \, dx$$ 8. **Calculate each integral separately:** - First integral: $$\int_0^2 e^x \, dx = [e^x]_0^2 = e^2 - 1$$ - Second integral (by integration by parts): Let $u = x$, $dv = e^x dx$ then $du = dx$, $v = e^x$. $$\int x e^x \, dx = x e^x - \int e^x \, dx = x e^x - e^x + C = e^x(x - 1) + C$$ Evaluate from 0 to 2: $$[e^x(x - 1)]_0^2 = e^2(2 - 1) - e^0(0 - 1) = e^2 - (-1) = e^2 + 1$$ 9. **Combine results:** $$\text{Area} = (e^2 - 1) - (e^2 + 1) = e^2 - 1 - e^2 - 1 = -2$$ 10. **Interpretation:** Area cannot be negative, so take absolute value: $$\text{Area} = 2$$ **Final answer:** $$\boxed{2}$$