Calculate the area of the figure bounded by the curve $y = \log x$, the straight line $x = 2$ and the X-axis.
The curve $y = \log x$ cuts the X-axis at the point $\left(1,0\right)$.
$\therefore \;$ The area between the curve $y = \log x$, the line $x = 2$ and the X-axis is:
$= \displaystyle \int_{1}^{2} y \; dx$
$= \displaystyle \int_{1}^{2} \log x \; dx$
$\begin{aligned}
\int \log x \; dx & = \log x \int 1 \; dx - \int \left\{\dfrac{d}{dx} \left(\log x\right) \times \int 1 \; dx \right\} dx \\\\
& \left[\text{Note: Integration by parts: } \right. \\
& \left. \int u \cdot v \; dx = u \int v \; dx - \int \left\{\dfrac{du}{dx} \times \int v \; dx \right\} dx\right] \\\\
& = x \log x - \int \dfrac{1}{x} \times x \; dx \\\\
& = x \log x - \int dx \\\\
& = x \log x - x + c
\end{aligned}$
$\begin{aligned}
\therefore \; \displaystyle \int_{1}^{2} \log x \; dx & = \left[x \log x\right]_{1}^{2} - \left[x\right]_{1}^{2} \\\\
& = \left[2 \times \log 2 - 1 \times \log 1\right] - \left[2 - 1\right] \\\\
& = 2 \log \left(2\right) - 1 \\\\
& = \log \left(4\right) - 1
\end{aligned}$
$\therefore \;$ Required area $= \log \left(4\right) - 1$ sq units