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Evaluate the Following Integral: 2 π ∫ 0 | Sin X | D X - Mathematics

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प्रश्न

Evaluate the following integral:

\[\int\limits_0^{2\pi} \left| \sin x \right| dx\]

 

योग
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उत्तर

\[\int_0^{2\pi} \left| \sin x \right| d x\]
\[\text{We know that}, \left| \sin x \right| = \begin{cases} - \sin x &,& \pi \leq x \leq 2\pi\\\sin x&,& 0 < x \leq \pi\end{cases}\]
\[ \therefore I = \int_0^{2\pi} \left| \sin x \right| dx\]
\[ \Rightarrow I = \int_0^\pi \sin x dx + \int_\pi^{2\pi} - \sin x dx\]
\[ \Rightarrow I = - \left[ \cos x \right]_0^\pi + \left[ \cos x \right]_\pi^{2\pi} \]
\[ \Rightarrow I = 1 + 1 + 1 - \left( - 1 \right)\]
\[ \Rightarrow I = 4\]

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अध्याय 20: Definite Integrals - Exercise 20.3 [पृष्ठ ५६]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 20 Definite Integrals
Exercise 20.3 | Q 12 | पृष्ठ ५६

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