1. **State the problem:** We have the curve $C$ defined by $y = x \ln x$ for $x > 0$, and the point $P(e, e)$ on $C$. The line $l$ is the normal to $C$ at $P$. The region $R$ is bounded by $C$, $l$, and the $x$-axis. We need to find the exact area of $R$ in the form $A e^2 + B$ where $A$ and $B$ are rational numbers. 2. **Find the derivative of $y = x \ln x$ to get the slope of the tangent at $P$:** $$y' = \frac{d}{dx}(x \ln x) = \ln x + 1$$ At $x = e$, $$y'(e) = \ln e + 1 = 1 + 1 = 2$$ So the slope of the tangent line at $P$ is 2. 3. **Find the slope of the normal line $l$ at $P$:** The normal slope is the negative reciprocal of the tangent slope: $$m_{normal} = -\frac{1}{2}$$ 4. **Equation of the normal line $l$ at $P(e, e)$:** Using point-slope form: $$y - e = -\frac{1}{2}(x - e)$$ Simplify: $$y = -\frac{1}{2}x + \frac{e}{2} + e = -\frac{1}{2}x + \frac{3e}{2}$$ 5. **Find the $x$-intercept of the normal line $l$ (where $y=0$):** $$0 = -\frac{1}{2}x + \frac{3e}{2}$$ Multiply both sides by 2: $$0 = -x + 3e$$ $$x = 3e$$ 6. **Find the $x$-intercept of the curve $C$ (where $y=0$):** $$x \ln x = 0$$ Since $x > 0$, $\ln x = 0 \Rightarrow x = 1$ 7. **Set up the area $R$ bounded by $C$, $l$, and the $x$-axis:** The region is between $x=1$ and $x=3e$. From $x=1$ to $x=e$, the upper boundary is the curve $y = x \ln x$. From $x=e$ to $x=3e$, the upper boundary is the normal line $y = -\frac{1}{2}x + \frac{3e}{2}$. 8. **Calculate the area under the curve from $1$ to $e$:** $$\int_1^e x \ln x \, dx$$ Use integration by parts: Let $u = \ln x$, $dv = x dx$. Then $du = \frac{1}{x} dx$, $v = \frac{x^2}{2}$. $$\int x \ln x \, dx = \frac{x^2}{2} \ln x - \int \frac{x^2}{2} \cdot \frac{1}{x} dx = \frac{x^2}{2} \ln x - \frac{1}{2} \int x \, dx = \frac{x^2}{2} \ln x - \frac{1}{2} \cdot \frac{x^2}{2} + C = \frac{x^2}{2} \ln x - \frac{x^2}{4} + C$$ Evaluate from 1 to $e$: $$\left[ \frac{x^2}{2} \ln x - \frac{x^2}{4} \right]_1^e = \left( \frac{e^2}{2} \cdot 1 - \frac{e^2}{4} \right) - \left( \frac{1}{2} \cdot 0 - \frac{1}{4} \right) = \left( \frac{e^2}{2} - \frac{e^2}{4} \right) - \left( 0 - \frac{1}{4} \right) = \frac{e^2}{4} + \frac{1}{4}$$ 9. **Calculate the area under the normal line from $e$ to $3e$:** $$\int_e^{3e} \left(-\frac{1}{2}x + \frac{3e}{2}\right) dx = \left[-\frac{1}{4} x^2 + \frac{3e}{2} x \right]_e^{3e}$$ Evaluate: At $x=3e$: $$-\frac{1}{4} (3e)^2 + \frac{3e}{2} (3e) = -\frac{9e^2}{4} + \frac{9e^2}{2} = -\frac{9e^2}{4} + \frac{18e^2}{4} = \frac{9e^2}{4}$$ At $x=e$: $$-\frac{1}{4} e^2 + \frac{3e}{2} e = -\frac{e^2}{4} + \frac{3e^2}{2} = -\frac{e^2}{4} + \frac{6e^2}{4} = \frac{5e^2}{4}$$ Subtract: $$\frac{9e^2}{4} - \frac{5e^2}{4} = \frac{4e^2}{4} = e^2$$ 10. **Total area $R$ is sum of both parts:** $$A = \frac{e^2}{4} + \frac{1}{4} + e^2 = \frac{e^2}{4} + e^2 + \frac{1}{4} = \frac{5e^2}{4} + \frac{1}{4}$$ Thus, the area is: $$A = \frac{5}{4} e^2 + \frac{1}{4}$$ **Answer:** $A = \frac{5}{4}$ and $B = \frac{1}{4}$, both rational numbers.