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प्रश्न
Evaluate the following integral:
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उत्तर
\[\int_0^3 \left| 3x - 1 \right| d x\]
\[\text{We know that}, \left| 3x - 1 \right| = \begin{cases} - \left( 3x - 1 \right)&,&0 \leq x \leq \frac{1}{3}\\\left( 3x - 1 \right)&,& \frac{1}{3} < x \leq 3\end{cases}\]
\[ \therefore I = = \int_0^\frac{1}{3} - \left( 3x + 1 \right) dx + \int_\frac{1}{3}^0 \left( 3x + 1 \right) dx\]
\[ \Rightarrow I = \left[ \frac{- 3 x^2}{2} - x \right]_0^\frac{1}{3} + \left[ \frac{3 x^2}{2} + x \right]_\frac{1}{3}^3 \]
\[ \Rightarrow I = - \frac{1}{6} + \frac{1}{3} - 0 + \frac{27}{2} + 3 - \frac{1}{6} - \frac{1}{3}\]
\[ \Rightarrow I = \frac{65}{6}\]
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