Question

Find area enclosed by $y = ex. lnx, y = \frac{ln x}{ex}$.

Answer 1

$\frac{e^2 - 5}{4e}$

Answer 2

To find the area enclosed by the two curves $y = ex \ln x$ and $y = \frac{\ln x}{ex}$, we first need to find their points of intersection.

1. Find the Intersection Points: Set the two equations equal to each other: $ex \ln x = \frac{\ln x}{ex}$ Multiply both sides by $ex$ (assuming $x \neq 0$): $e^2 x^2 \ln x = \ln x$ $e^2 x^2 \ln x - \ln x = 0$ $\ln x (e^2 x^2 - 1) = 0$

This equation gives two possibilities:

• $\ln x = 0 \implies x = e^0 = 1$.
• $e^2 x^2 - 1 = 0 \implies e^2 x^2 = 1 \implies x^2 = \frac{1}{e^2} \implies x = \pm \frac{1}{e}$. Since $\ln x$ is defined only for $x > 0$, we take $x = \frac{1}{e}$.

So, the intersection points are $x = \frac{1}{e}$ and $x = 1$.

2. Determine Which Curve is Above the Other: Let $f(x) = ex \ln x$ and $g(x) = \frac{\ln x}{ex}$. We need to determine which function is greater in the interval $(\frac{1}{e}, 1)$. Consider the difference $g(x) - f(x)$: $g(x) - f(x) = \frac{\ln x}{ex} - ex \ln x = \ln x \left( \frac{1}{ex} - ex \right) = \ln x \left( \frac{1 - e^2 x^2}{ex} \right)$.

In the interval $x \in (\frac{1}{e}, 1)$:

• $\ln x$ is negative (since $x < 1$).
• Since $x > \frac{1}{e}$, we have $x^2 > \frac{1}{e^2}$, which implies $e^2 x^2 > 1$. Therefore, $1 - e^2 x^2$ is negative.
• $ex$ is positive.

So, the term $\ln x \left( \frac{1 - e^2 x^2}{ex} \right)$ is (negative) $\times$ (negative) $\times$ (positive) = positive. This means $g(x) - f(x) > 0$, so $g(x)$ is the upper curve and $f(x)$ is the lower curve in the interval $(\frac{1}{e}, 1)$.

3. Set up the Definite Integral for the Area: The area $A$ enclosed by the curves is given by: $A = \int_{1/e}^{1} (g(x) - f(x)) dx = \int_{1/e}^{1} \left( \frac{\ln x}{ex} - ex \ln x \right) dx$

4. Evaluate the Integral: We integrate term by term.

• Integral of the first term: $\int \frac{\ln x}{ex} dx = \frac{1}{e} \int \frac{\ln x}{x} dx$. Let $u = \ln x$, then $du = \frac{1}{x} dx$. $\frac{1}{e} \int u du = \frac{1}{e} \frac{u^2}{2} = \frac{(\ln x)^2}{2e}$.

• Integral of the second term: $\int ex \ln x dx$. Use integration by parts: $\int P Q' dx = PQ - \int P' Q dx$. Let $P = \ln x$ and $Q' = ex$. Then $P' = \frac{1}{x}$ and $Q = \frac{ex^2}{2}$. $\int ex \ln x dx = (\ln x) \left( \frac{ex^2}{2} \right) - \int \left( \frac{1}{x} \right) \left( \frac{ex^2}{2} \right) dx$ $= \frac{ex^2 \ln x}{2} - \int \frac{ex}{2} dx$ $= \frac{ex^2 \ln x}{2} - \frac{ex^2}{4}$.

Now, combine these results to find the indefinite integral of $g(x) - f(x)$: $\int \left( \frac{\ln x}{ex} - ex \ln x \right) dx = \frac{(\ln x)^2}{2e} - \left( \frac{ex^2 \ln x}{2} - \frac{ex^2}{4} \right)$ $= \frac{(\ln x)^2}{2e} - \frac{ex^2 \ln x}{2} + \frac{ex^2}{4}$.

Now, evaluate this definite integral from $x = \frac{1}{e}$ to $x = 1$: $A = \left[ \frac{(\ln x)^2}{2e} - \frac{ex^2 \ln x}{2} + \frac{ex^2}{4} \right]_{1/e}^{1}$

• At the upper limit $x=1$: $\frac{(\ln 1)^2}{2e} - \frac{e(1)^2 \ln 1}{2} + \frac{e(1)^2}{4} = \frac{0}{2e} - \frac{e \cdot 0}{2} + \frac{e}{4} = \frac{e}{4}$.

• At the lower limit $x=\frac{1}{e}$: Note that $\ln(\frac{1}{e}) = \ln(e^{-1}) = -1$. $\frac{(\ln(\frac{1}{e}))^2}{2e} - \frac{e(\frac{1}{e})^2 \ln(\frac{1}{e})}{2} + \frac{e(\frac{1}{e})^2}{4}$ $= \frac{(-1)^2}{2e} - \frac{e \cdot \frac{1}{e^2} \cdot (-1)}{2} + \frac{e \cdot \frac{1}{e^2}}{4}$ $= \frac{1}{2e} - \frac{\frac{1}{e} \cdot (-1)}{2} + \frac{\frac{1}{e}}{4}$ $= \frac{1}{2e} + \frac{1}{2e} + \frac{1}{4e}$ $= \frac{2}{2e} + \frac{1}{4e} = \frac{1}{e} + \frac{1}{4e} = \frac{4+1}{4e} = \frac{5}{4e}$.

Finally, subtract the value at the lower limit from the value at the upper limit: $A = \frac{e}{4} - \frac{5}{4e} = \frac{e^2}{4e} - \frac{5}{4e} = \frac{e^2 - 5}{4e}$.

The area enclosed by the curves is $\frac{e^2 - 5}{4e}$.