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JEE Main 202226 JulyEvening Shift
Exam
JEE Main
Year
2022
Date
26 July
Shift
Evening Shift
Subject
Mathematics
Chapter
Definite Integration

JEE Main 2022 · 26 July · Evening Shift · Mathematics · Definite Integration

∫ _ 0 ^ 20π (| x| + | x|) ^ 2 dx is equal to:

∫020π(∣sin⁡x∣+∣cos⁡x∣)2dx{∫}_{0}^{20\pi}{\left(\left|\sin x\right| + \left|\cos x\right|\right)}^{2}dx is equal to:

Options

  1. A10(π+4)10\left(\pi + 4\right)
  2. B10(π+2)10\left(\pi + 2\right)
  3. C20(π−2)20\left(\pi - 2\right)
  4. D20(π+2)20\left(\pi + 2\right)

Correct answer

D. 20(π+2)20\left(\pi + 2\right)

Step-by-step solution

Let I=∫020π(∣sin⁡x∣+∣cos⁡x∣)2dxI = {∫}_{0}^{20\pi}{\left(\left|\sin x\right| + \left|\cos x\right|\right)}^{2}dx

We know that the period of \left|\sin x\right| + \left|\cos x\right|isis\dfrac{\pi}{2}</p><p>i.e.</p><p>i.e.I = 40{∫}_{0}^{\dfrac{\pi}{2}}{\left(\sin x + \cos x\right)}^{2}dx</p><p></p><p>I = 40{∫}_{0}^{\dfrac{\pi}{2}}\left(1 + \sin 2x\right)dx</p><p></p><p>= 40{\left(x - \dfrac{\cos 2x}{2}\right)}_{0}^{\dfrac{\pi}{2}} = 40\left(\dfrac{\pi}{2} - \dfrac{cos \pi}{2} + \dfrac{\cos 0}{2}\right)</p><p></p><p>I = 20 \left[\pi + 2\right]

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