Q86 / 150 · MHT CET 2026 (15 April Shift 2) · Mathematics
$\int\limits_{0}^{\pi} |\sin 2x|\, \text{d}x =$
- A$0$
- B$1$
- Correct answerC$2$
- D$-1$
Solution
Let $I = \int_{0}^{\pi} |\sin 2x| \, dx$
Using the property $\int_{0}^{2a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$ if $f(2a - x) = f(x)$, we check $f(\pi - x) = |\sin(2\pi - 2x)| = |-\sin 2x| = |\sin 2x| = f(x)$.
Therefore, $I = 2 \int_{0}^{\pi/2} |\sin 2x| \, dx$
In the interval $[0, \pi/2]$, $2x \in [0, \pi]$, which means $\sin 2x \ge 0$.
$I = 2 \int_{0}^{\pi/2} \sin 2x \, dx$
$I = 2 \left[ -\dfrac{\cos 2x}{2} \right]_{0}^{\pi/2}$
$I = -(\cos \pi - \cos 0)$
$I = -(-1 - 1) = 2$
Answer: $2$
Using the property $\int_{0}^{2a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$ if $f(2a - x) = f(x)$, we check $f(\pi - x) = |\sin(2\pi - 2x)| = |-\sin 2x| = |\sin 2x| = f(x)$.
Therefore, $I = 2 \int_{0}^{\pi/2} |\sin 2x| \, dx$
In the interval $[0, \pi/2]$, $2x \in [0, \pi]$, which means $\sin 2x \ge 0$.
$I = 2 \int_{0}^{\pi/2} \sin 2x \, dx$
$I = 2 \left[ -\dfrac{\cos 2x}{2} \right]_{0}^{\pi/2}$
$I = -(\cos \pi - \cos 0)$
$I = -(-1 - 1) = 2$
Answer: $2$
Hence, the correct answer is option: 'C'.