The area bounded by the curve \( y = \log x , y = 0 \), and \( x = e \), is

1

The correct answer is Option (3) → 1

Given: Find the area bounded by the curve $y = \log x$, the line $y = 0$ (x-axis), and $x = e$.

Limits: From $x = 1$ to $x = e$ (since $\log 1 = 0$ and $x = e$ is the upper bound)

Required Area:

$\int_{1}^{e} \log x \, dx$

Use integration by parts:

Let $u = \log x$, $dv = dx$ ⟹ $du = \frac{1}{x} dx$, $v = x$

$\int \log x \, dx = x \log x - x + C$

So,

$\int_{1}^{e} \log x \, dx = [x \log x - x]_{1}^{e}$

$= (e \cdot \log e - e) - (1 \cdot \log 1 - 1)$

$= (e \cdot 1 - e) - (0 - 1)$

$= (e - e) - (-1) = 0 + 1 = 1$