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Evaluate the Following Integral: 1 ∫ − 1 | 2 X + 1 | D X - Mathematics

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प्रश्न

Evaluate the following integral:

\[\int\limits_{- 1}^1 \left| 2x + 1 \right| dx\]
योग
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उत्तर

\[\int_{- 1}^1 \left| 2x + 1 \right| d x\]
\[We\ know\ that\, \left| 2x + 1 \right| = \begin{cases} - \left( 2x + 1 \right) &, &- 1 \leq x \leq - \frac{1}{2} \\\left( 2x + 1 \right) &, &- \frac{1}{2} < x \leq 1\end{cases}\]
\[ \therefore I = \int_{- 1}^\frac{- 1}{2} - \left( 2x + 1 \right) d x + \int_{- \frac{1}{2}}^1 \left( 2x + 1 \right) d x\]
\[ \Rightarrow I = - \left[ x^2 + x \right]_{- 1}^\frac{- 1}{2} + \left[ x^2 + x \right]_{- \frac{1}{2}}^1 \]
\[ \Rightarrow I = - \frac{1}{4} + \frac{1}{2} + 1 - 1 + 1 + 1 - \frac{1}{4} + \frac{1}{2}\]
\[ \Rightarrow I = \frac{5}{2}\]

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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 4 | पृष्ठ ५६

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