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BITSAT 2011
Exam
BITSAT
Year
2011
Subject
Mathematics
Chapter
Definite Integration

BITSAT 2011 · Mathematics · Definite Integration

Find the value of (∫_0^ 4 π | x| d x)

Find the value of (\int_0^{4 \pi}|\sin x| d x)

Options

  1. A8
  2. B6
  3. C4
  4. D2

Correct answer

A. 8

Step-by-step solution

We know that (|\sin x|) is a periodic function of (\pi) (\text{ Hence } ∫04π∣sin⁡x∣dx=4∫0π∣\sinx∣dx=4∫0π\sinxdx=4[−cos⁡x]0π=8\begin{array}{r} \int_0^{4 \pi}|\sin x| d x=4 \int_0^\pi|\sinx| d x=4 \int_0^\pi \sinx d x \\ =4[-\cos x]_0^\pi=8 \end{array})
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