${\int }_1^4|x-2|dx\mathrm{is}\mathrm{equal}\mathrm{to}$

The correct option is:
$\frac{5}{2}$

Step-by-Step Explanation:

We are required to evaluate the definite integral:
$I={\int }_1^4|x-2|dx$

First, let us analyze the behavior of the integrand, which contains an absolute value term $|x-2|$. By definition, the absolute value function is defined as:
$|x-2|=\{\begin{array}{l}x-2\text{if }x\ge 2\\ -(x-2)=2-x\text{if }x<2\end{array}$

The interval of integration is from 1 to 4. Since the definition of the integrand changes at $x=2$, we can split the integration interval $[1,4]$ into two sub-intervals: $[1,2]$ and $[2,4]$. This allows us to write:
$I={\int }_1^2|x-2|dx+{\int }_2^4|x-2|dx$

Substituting the corresponding expressions for the absolute value in each interval:
$I={\int }_1^2(2-x)dx+{\int }_2^4(x-2)dx$

Now, let us evaluate each integral separately.

Part 1: Evaluating the first integral
$I_1={\int }_1^2(2-x)dx$
Using the power rule for integration, we get:
$I_1={[2x-\frac{x^2}{2}]}_1^2$
Applying the upper and lower limits:
$I_1=\left(2\right(2)-\frac{2^2}{2})-\left(2\right(1)-\frac{1^2}{2})$
Simplifying terms:
$I_1=(4-2)-(2-\frac{1}{2})$
$I_1=2-\frac{3}{2}=\frac{1}{2}$

Part 2: Evaluating the second integral
$I_2={\int }_2^4(x-2)dx$
Integrating the terms:
$I_2={[\frac{x^2}{2}-2x]}_2^4$
Applying the upper and lower limits:
$I_2=(\frac{4^2}{2}-2(4\left)\right)-(\frac{2^2}{2}-2(2\left)\right)$
Simplifying terms:
$I_2=(8-8)-(2-4)$
$I_2=0-(-2)=2$

Summing the parts:
Combining the values of both integrals to find the final value of $I$:
$I=I_1+I_2$
$I=\frac{1}{2}+2=\frac{5}{2}$