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Question
\[\int\limits_1^3 \left| x^2 - 2x \right| dx\]
Solution
We have,
\[\left| x^2 - 2x \right| = \begin{cases}- \left( x^2 - 2x \right),& 1 \leq x \leq 2\\ x^2 - 2x,& 2 \leq x \leq 3\end{cases}\]
\[ \therefore \int_1^3 \left| x^2 - 2x \right| d x\]
\[ = \int_1^2 - \left( x^2 - 2x \right) dx + \int_2^3 \left( x^2 - 2x \right) dx\]
\[ = \left[ - \frac{x^3}{3} + x^2 \right]_1^2 + \left[ \frac{x^3}{3} - x^2 \right]_2^3 \]
\[ = \frac{- 8}{3} + 4 + \frac{1}{3} - 1 + 9 - 9 - \frac{8}{3} + 4\]
= 2
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