Determine the value of \(\int_0 ^{\frac{\pi}{2}
}(-2cos x)dx\)
}(-2cos x)dx\)
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option A
Explanation:
\(\int_0 ^{\frac{\pi}{2}}(-2cos x)dx = [-2sin x + c]_0 ^{\frac{\pi}{2}}\\
=(-2sin\frac{\pi}{2}+c+2sin0-c)\\
=-2sin90+c+2sin0-c\\
=-2(1)+2(0)\\
=-2\)
\(\int_0 ^{\frac{\pi}{2}}(-2cos x)dx = [-2sin x + c]_0 ^{\frac{\pi}{2}}\\
=(-2sin\frac{\pi}{2}+c+2sin0-c)\\
=-2sin90+c+2sin0-c\\
=-2(1)+2(0)\\
=-2\)