Evaluate \(\int^{\frac{\pi}{2}} _{\frac{-\pi}{2}} cos x dx\)

  • A zero
  • B 1
  • C 2
  • D 3

The correct answer is C. 2

The indefinite integral of cos x is sin x + C, where C is the constant of integration. To evaluate the definite integral \(\int^{\frac{\pi}{2}} _{\frac{-\pi}{2}} cos x dx\), we can use the Fundamental Theorem of Calculus, which states that if F(x) is an antiderivative of f(x) on an interval [a, b], then \(\int^b_a f(x) dx = F(b) - F(a)\).

In this case, we can take F(x) = sin x, which is an antiderivative of f(x) = cos x. Then, we have:

\(\int^{\frac{\pi}{2}} _{\frac{-\pi}{2}} cos x dx = sin(\frac{\pi}{2}) - sin(\frac{-\pi}{2}) = 1 - (-1) = 2\)

So, the correct answer is 2

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