$\text { The area of the region (in square units) bounded by the line } y+3=x ; x=1 \text { and } x=5 \text { is }$

The region is bounded by the line $y = x - 3$, the vertical lines $x = 1$ and $x = 5$, and the x-axis ($y = 0$).

The area $A$ is given by the definite integral of the absolute value of the function $f(x) = x - 3$ from $x = 1$ to $x = 5$.

$A = \int_{1}^{5} |x - 3| dx$

The function $x - 3$ changes sign at $x = 3$. Thus, the integral is split as follows:

$A = \int_{1}^{3} -(x - 3) dx + \int_{3}^{5} (x - 3) dx$

Evaluating the first integral:

$\int_{1}^{3} (3 - x) dx = [3x - \dfrac{x^2}{2}]_{1}^{3} = (9 - \dfrac{9}{2}) - (3 - \dfrac{1}{2}) = 4.5 - 2.5 = 2$

Evaluating the second integral:

$\int_{3}^{5} (x - 3) dx = [\dfrac{x^2}{2} - 3x]_{3}^{5} = (\dfrac{25}{2} - 15) - (\dfrac{9}{2} - 9) = (12.5 - 15) - (4.5 - 9) = -2.5 - (-4.5) = 2$

The total area is $2 + 2 = 4$.

Answer: 4