The area of the region bounded by the curves $\mathrm{y}=|x-4|, x=3$ and $x=5$, and the X -axis is

The area under $y=|x-4|$ between $x=3$ and $x=5$ is given by the integral $\int_{3}^{5} |x-4| dx$.

The absolute value function splits at $x=4$, requiring evaluation in two parts:

$\int_{3}^{5} |x-4| dx = \int_{3}^{4} (4-x) dx + \int_{4}^{5} (x-4) dx$

Evaluating the first integral:

$\int_{3}^{4} (4-x) dx = \left[4x - \frac{x^2}{2}\right]_{3}^{4} = (16 - 8) - (12 - 4.5) = 8 - 7.5 = 0.5$

The second integral yields:

$\int_{4}^{5} (x-4) dx = \left[\frac{x^2}{2} - 4x\right]_{4}^{5} = (12.5 - 20) - (8 - 16) = (-7.5) - (-8) = 0.5$

Summing both contributions gives the total area:
$0.5 + 0.5 = 1$

The area is 1 square unit.