JAMB - Mathematics (2005 - No. 33)
Evaluate \(\int_0^{\frac{\pi}{2}}sin2xdx\)
1
zero
-1/2
-1
स्पष्टीकरण
\(\int_0^{\frac{\pi}{2}}\)sin 2x dx = [-1/2cos 2x + C]\(_0^{\frac{\pi}{2}}\)
=[-1/2 cos 2 * π/2 + C] - [-1/2 cos 2 * 0]
= [-1/2 cos π] - [-1/2 cos 0]
= [-1/2x - 1] - [-1/2 * 1]
= 1/2 -(-1/2) = 1/2 + 1/2 = 1
=[-1/2 cos 2 * π/2 + C] - [-1/2 cos 2 * 0]
= [-1/2 cos π] - [-1/2 cos 0]
= [-1/2x - 1] - [-1/2 * 1]
= 1/2 -(-1/2) = 1/2 + 1/2 = 1