Q100 / 180 · COMEDK 2025 (10 May Shift 3) · Mathematics
$\int_{-2}^2 \dfrac{|x-3|}{x-3} d x=$
- Correct answerA$-4$
- B$-2$
- C$0$
- D$2$
Solution
The integrand is $f(x) = \dfrac{|x-3|}{x-3}$.
Recall the definition of the absolute value function: $|x-3| = x-3$ if $x-3 > 0$ (i.e., $x > 3$) and $|x-3| = -(x-3)$ if $x-3 < 0$ (i.e., $x < 3$).
For the interval $[-2, 2]$, we have $x < 3$ for all $x$ in the domain of integration.
Therefore, for all $x \in [-2, 2]$, $|x-3| = -(x-3)$.
Substituting this into the integrand, we get $f(x) = \dfrac{-(x-3)}{x-3} = -1$.
The integral becomes $\int_{-2}^{2} (-1) dx$.
Evaluating the definite integral: $[-x]_{-2}^{2} = -(2) - (-(-2)) = -2 - 2 = -4$.
Answer: $-4$
Recall the definition of the absolute value function: $|x-3| = x-3$ if $x-3 > 0$ (i.e., $x > 3$) and $|x-3| = -(x-3)$ if $x-3 < 0$ (i.e., $x < 3$).
For the interval $[-2, 2]$, we have $x < 3$ for all $x$ in the domain of integration.
Therefore, for all $x \in [-2, 2]$, $|x-3| = -(x-3)$.
Substituting this into the integrand, we get $f(x) = \dfrac{-(x-3)}{x-3} = -1$.
The integral becomes $\int_{-2}^{2} (-1) dx$.
Evaluating the definite integral: $[-x]_{-2}^{2} = -(2) - (-(-2)) = -2 - 2 = -4$.
Answer: $-4$
Hence, the correct answer is option: 'A'.