Question

Evaluate the following integral:

\[\int\limits_1^2 \left| x - 3 \right| dx\]

Answer 1

\[\int_1^2 \left| x - 3 \right| d x\]
\[\text{We know that}, \left| x + 1 \right| = \begin{cases} - \left( x + 1 \right) &,& 1 \leq x \leq 3\\\left( x + 1 \right)&,& x > 3\end{cases}\]
\[ \therefore I = \int_1^2 \left| x - 3 \right| d x\]
\[ \Rightarrow I = \int_1^2 - \left( x - 3 \right) dx\]
\[ \Rightarrow I = \left[ \frac{- x^2}{2} - 3x \right]_1^2 \]
\[ \Rightarrow I = - 2 - 6 + \frac{1}{2} + 3\]

\[ \Rightarrow I = - 5 + \frac{1}{2}\]

\[ \Rightarrow I = \frac{-10 + 1}{2}\]
\[ \Rightarrow I = -\frac{9}{2}\]