JAMB - Mathematics (2008 - No. 35)
Evaluate \(\int^{\frac{\pi}{2}} _{\frac{-\pi}{2}} cos x dx\)
zero
1
2
3
คำอธิบาย
\(\int^{\frac{\pi}{2}} _{\frac{-\pi}{2}} cos x dx = [sinx]^{\frac{\pi}{2}} _{\frac{-\pi}{2}}\\
=sin\frac{\pi}{2} - sin\frac{-\pi}{2}\)
= sin90 – sin-90
= sin90 – sin270
= 1 – (-1)
= 1+1
= 2
=sin\frac{\pi}{2} - sin\frac{-\pi}{2}\)
= sin90 – sin-90
= sin90 – sin270
= 1 – (-1)
= 1+1
= 2