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Question
\[\int\limits_1^3 \left| x^2 - 4 \right| dx\]
Solution
\[\int_1^3 \left| x^2 - 4 \right| d x\]
\[ = \int_1^2 - \left( x^2 - 4 \right) dx + \int_2^3 \left( x^2 - 4 \right) dx\]
\[ = \left[ - \frac{x^3}{3} + 4x \right]_1^2 + \left[ \frac{x^3}{3} - 4x \right]_2^3 \]
\[ = \frac{- 8}{3} + 8 + \frac{1}{3} - 4 + 9 - 12 - \frac{8}{3} + 8\]
\[ = 4\]
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