Integrals
NCERT Class 12
CBSE
Grade 12
Question:
The value of $\int_0^{\pi/2} \sin^2 x dx$ is:
(a) $\dfrac{\pi}{4}$
(b) $\dfrac{\pi}{2}$
(c) $\dfrac{\pi}{8}$
(d) $\pi$
Step-by-Step Solution
$\left[ \dfrac{x}{2} - \dfrac{\sin 2x}{4} \right]_0^{\pi/2} = \dfrac{\pi}{4}$. [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Evaluating definite integral $= \pi/4$: 1.0 Mark
Correct Answer: $\dfrac{\pi}{4}$
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