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Evaluate: \(\int^{z}_{0}(sin x - cos x) dx \hspace{1mm} Where\hspace{1mm}letter\hspace{1mm}z = \frac{\pi}{4}. (\pi = pi)\)

Mathematics
JAMB 1999

Evaluate: \(\int^{z}_{0}(sin x - cos x) dx \hspace{1mm}

Where\hspace{1mm}letter\hspace{1mm}z = \frac{\pi}{4}. (\pi = pi)\)

  • A. \(\sqrt{2 +1}\)
  • B. \(\sqrt{2 }-1\)
  • C. \(-\sqrt{2 }-1\)
  • D. \(1-\sqrt{2}\)
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Correct Answer: Option D
Explanation

\(\int_{0}^{z} (\sin x - \cos x) \, dx\)

where \( z = \frac{\pi}{4} \), we have:

\(\int_{0}^{\frac{\pi}{4}} (\sin x - \cos x) \, dx = \left[-\cos x - \sin x\right]_{0}^{\frac{\pi}{4}}\)

Evaluating at the bounds gives:

\(-\sqrt{2} - (-1) = -\sqrt{2} + 1\)

Thus, the value of the integral is:

\(\int_{0}^{\frac{\pi}{4}} (\sin x - \cos x) \, dx = 1 - \sqrt{2}\)


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