The area of the region bounded by the and the lines x = 2 and x = 3

The correct option is 7/2 sq. unit.

Problem Clarification:
The given problem is to find the area of the region bounded by the curve
$y=x+1$
the x-axis, and the vertical lines
$x=2$
and
$x=3$.

Step-by-Step Derivation:

Step 1: Set up the definite integral for the area
The area A of a region bounded by a curve
$y=f\left(x\right)$
the x-axis, and the lines
$x=a$
and
$x=b$
is given by the definite integral:
$A={\int }_a^{b}f\left(x\right) dx$
Substituting
$f\left(x\right)=x+1$, $a=2$, and $b=3$,
we get:
$A={\int }_2^3(x+1) dx$

Step 2: Find the antiderivative
Integrating each term of the integrand with respect to x using the power rule:
$\int (x+1) dx=\frac{x^2}{2}+x$
Applying the limits of integration from 2 to 3, we have:
$A={\left[\frac{x^2}{2}+x\right]}_2^3$

Step 3: Evaluate the limits
First, substitute the upper limit
$x=3$:
$F_{\text{upper}}=\frac{3^2}{2}+3=\frac{9}{2}+3=\frac{9}{2}+\frac{6}{2}=\frac{15}{2}$
Next, substitute the lower limit
$x=2$:
$F_{\text{lower}}=\frac{2^2}{2}+2=\frac{4}{2}+2=2+2=4$

Step 4: Subtract the evaluated lower limit from the upper limit
Subtracting the lower limit value from the upper limit value:
$A=F_{\text{upper}}-F_{\text{lower}}$
$A=\frac{15}{2}-4$
$A=\frac{15}{2}-\frac{8}{2}=\frac{7}{2}$
Thus, the area of the region is indeed 7/2 sq. unit.