NIMCET 2022 — Mathematics PYQ

To evaluate the integral ${\int }_a^b(|x-a|+|x-b|)dx$ where a < b, we analyze the behavior of the modulus functions within the interval $[a,b]$.

1. Splitting the integral:

Since $a\le x\le b$:

• For $|x-a|$: Since $x\ge a$, $|x-a|=x-a$.

• For $|x-b|$: Since $x\le b$, $|x-b|=-(x-b)=b-x$.

Substituting these into the integral:

$$I={\int }_a^b((x-a)+(b-x))dx$$

2. Simplifying the integrand:

$$I={\int }_a^b(x-a+b-x)dx$$

$$I={\int }_a^b(b-a)dx$$

3. Evaluating the definite integral:

Since $(b-a)$ is a constant with respect to $x$:

$$I=(b-a){\int }_a^{b}1dx$$

$$I=(b-a)[x{]}_a^b$$

$$I=(b-a)(b-a)$$

$$I=(b-a{)}^2$$

Conclusion:

The value of the definite integral is $(b-a{)}^2$. Therefore, the correct option is (D).