Find the area enclosed by the curve y = (ln 2 + ln

Find the area enclosed by the curve $y = \frac{\ln 2 + \ln\sqrt{x}}{\sqrt{x}}$, the $x$-axis, and the ordinate (vertical line) at the maximum point.

We will simplify with $u=\sqrt x$, locate the maximum and the axis-crossing, then integrate using the same substitution.

Step 1: Simplify the function

\[y = \frac{\ln 2 + \ln\sqrt{x}}{\sqrt{x}} = \frac{\ln 2 + \frac{1}{2}\ln x}{\sqrt{x}} = \frac{\ln(2\sqrt{x})}{\sqrt{x}}\]

Step 2: Find the maximum

Let $u = \sqrt{x}$, so $x = u^2$ and:

\[y = \frac{\ln 2 + \ln u}{u}\]

Taking the derivative with respect to $u$:

\[\frac{dy}{du} = \frac{\frac{1}{u} \cdot u - (\ln 2 + \ln u) \cdot 1}{u^2} = \frac{1 - \ln 2 - \ln u}{u^2}\]

Setting this equal to zero:

\[1 - \ln 2 - \ln u = 0 \implies \ln u = 1 - \ln 2 = \ln\frac{e}{2} \implies u = \frac{e}{2}\]

Since $u = \sqrt{x}$:

\[\sqrt{x} = \frac{e}{2} \implies x = \frac{e^2}{4}\]

Verify this is a maximum: for $u<\frac{e}{2}$, $\frac{dy}{du}>0$ (increasing); for $u>\frac{e}{2}$, $\frac{dy}{du}<0$ (decreasing). So $x=\frac{e^2}{4}$ gives a maximum. ✓

Step 3: Find the maximum value

At $x = \frac{e^2}{4}$ (so $u=\frac e2$):

\[y_{\max} = \frac{\ln 2 + \ln\frac{e}{2}}{\frac{e}{2}} = \frac{\ln 2 + \ln e - \ln 2}{\frac{e}{2}} = \frac{1}{\frac{e}{2}} = \frac{2}{e}\]

Step 4: Determine the region

The curve intersects the $x$-axis where $y = 0$:

\[\ln(2\sqrt{x}) = 0 \implies 2\sqrt{x} = 1 \implies \sqrt{x} = \frac{1}{2} \implies x = \frac{1}{4}\]

The region is bounded by the curve from $x = \frac{1}{4}$ to $x = \frac{e^2}{4}$, the $x$-axis, and the vertical line at $x = \frac{e^2}{4}$.

Step 5: Set up the integral

\[A = \int_{1/4}^{e^2/4} \frac{\ln 2 + \ln\sqrt{x}}{\sqrt{x}} \, dx\]

Let $u = \sqrt{x}$, so $x = u^2$ and $dx = 2u \, du$. When $x = \frac{1}{4}$: $u = \frac{1}{2}$. When $x = \frac{e^2}{4}$: $u = \frac{e}{2}$.

\[A = \int_{1/2}^{e/2} \frac{\ln 2 + \ln u}{u} \cdot 2u \, du = 2\int_{1/2}^{e/2} (\ln 2 + \ln u) \, du\]

Step 6: Evaluate the integral

First piece:

\[2\ln 2 \int_{1/2}^{e/2} du = 2\ln 2\left[\frac{e}{2} - \frac{1}{2}\right] = (e-1)\ln 2\]

Second piece: using $\int \ln u \, du = u\ln u - u + C$:

\[2\int_{1/2}^{e/2} \ln u \, du = 2[u\ln u - u]_{1/2}^{e/2}\]

At $u=\frac e2$: $\frac{e}{2}\ln\frac{e}{2} - \frac{e}{2} = \frac{e}{2}(1-\ln2) - \frac{e}{2} = -\frac{e\ln2}{2}$. At $u=\frac12$: $\frac12\ln\frac12-\frac12 = -\frac{\ln2}{2}-\frac12$.

\[2\left[-\frac{e\ln 2}{2} - \left(-\frac{\ln 2}{2} - \frac{1}{2}\right)\right] = -e\ln 2 + \ln 2 + 1 = (1-e)\ln 2 + 1\]

Step 7: Total area

\[A = (e-1)\ln 2 + (1-e)\ln 2 + 1 = 1\]

\[\boxed{A = 1 \text{ square unit}}\]

Summary

\[\boxed{A = 1 \text{ square unit}}\]

Remarkable result: the two $\ln2$-dependent pieces cancel exactly, leaving the enclosed area independent of the constant $2$ inside the logarithm — exactly 1 square unit, regardless of the details of the curve's shape.