Determine the value of (int_0 ^{frac{pi}{2}
}(-2cos x)dx)
}(-2cos x)dx)
-
A.
-2 -
B.
–1/2 -
C.
-3 -
D.
–3/2
Correct Answer: Option A
Explanation
(int_0 ^{frac{pi}{2}}(-2cos x)dx = [-2sin x + c]_0 ^{frac{pi}{2}}\
=(-2sinfrac{pi}{2}+c+2sin0-c)\
=-2sin90+c+2sin0-c\
=-2(1)+2(0)\
=-2)