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NDA 2019
Exam
NDA
Year
2019
Subject
Mathematics
Chapter
Definite Integration

NDA 2019 · Mathematics · Definite Integration

∫_0^ (π)/(2) | x- x| d x is equal to

∫0π2∣sin⁡x−cos⁡x∣dx\int_0^{\frac{\pi}{2}}|\sin x-\cos x| d x is equal to

Options

  1. A0
  2. B2(2−1)2(\sqrt{2}-1)
  3. C222 \sqrt{2}
  4. D2(2+1)2(\sqrt{2}+1)

Correct answer

B. 2(2−1)2(\sqrt{2}-1)

Step-by-step solution

∫0π2∣sin⁡x−cos⁡x∣dx\int_0^{\frac{\pi}{2}}|\sin x-\cos x| d x=∫0π4∣(sin⁡x−cos⁡x)∣dx+∫π4π2∣(sin⁡x−cos⁡x)∣dx=∫0π4(cos⁡x−sin⁡x)⋅dx+∫π4π2(sin⁡x−cos⁡x)⋅dx=(sin⁡x+cos⁡x)0π4+(−cos⁡x−sin⁡x)π2π2=22−2=2(2−1)\begin{aligned} & =\int_0^{\frac{\pi}{4}}|(\sin x-\cos x)| d x+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}|(\sin x-\cos x)| d x \\ & =\int_0^{\frac{\pi}{4}}(\cos x-\sin x) \cdot d x+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}(\sin x-\cos x) \cdot d x \\ & =(\sin x+\cos x)_0^{\frac{\pi}{4}}+(-\cos x-\sin x)_{\frac{\pi}{2}}^{\frac{\pi}{2}} \\ & =2 \sqrt{2}-2=2(\sqrt{2}-1)\end{aligned}$$
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