The curve $y = \ln x$ cuts the $x$-axis at $P$. The area, $A$ units$^2$, is enclosed by the curve, the $x$-axis and the line $x = e^2$. Explain why $e^2 - 1 < \int_{1}^{e^2} y \, dx < 2(e^2 - 1)$.

1. Find the coordinates of the endpoints of the region

At the $x$-axis, $y = 0 \implies \ln x = 0 \implies x = 1$. So $P$ is $(1, 0)$.
At $x = e^2$, $y = \ln(e^2) = 2\ln e = 2$.
The endpoints of the curve segment are $(1, 0)$ and $(e^2, 2)$.

2. Explain the upper bound using a bounding rectangle

The integral $\int_{1}^{e^2} y \, dx$ represents the exact area under the curve $y = \ln x$ from $x = 1$ to $x = e^2$.
The region is entirely contained within a rectangle of width $(e^2 - 1)$ and height $2$.
Area of this rectangle $= 2(e^2 - 1)$.
Since the curve is strictly increasing and lies below the line $y = 2$ for $1 \leqslant x < e^2$, the area under the curve is strictly less than the area of the rectangle:
$\int_{1}^{e^2} y \, dx < 2(e^2 - 1)$

3. Explain the lower bound using an inscribed triangle

Consider the triangle formed by the points $(1, 0)$, $(e^2, 0)$, and $(e^2, 2)$.
Area of this triangle $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}(e^2 - 1)(2) = e^2 - 1$.
The curve $y = \ln x$ is concave downwards (its gradient decreases as $x$ increases), so the curve lies above the chord connecting $(1, 0)$ and $(e^2, 2)$.
Therefore, the area under the curve is strictly greater than the area of the triangle:
$\int_{1}^{e^2} y \, dx > e^2 - 1$

4. Combine the inequalities

Combining the two results, we have:
$e^2 - 1 < \int_{1}^{e^2} y \, dx < 2(e^2 - 1)$