Question

Evaluate the following integral:

\[\int\limits_0^2 \left| x^2 - 3x + 2 \right| dx\]

 

Answer 1

\[\int_0^2 \left| x^2 - 3x + 2 \right| d x\]
\[\text{We know that}, \left| x^2 - 3x + 2 \right| = \begin{cases} - \left( x^2 - 3x + 2 \right)&, &\left( x - 1 \right)\left( x - 2 \right) \leq 0 \text{ or}, 1 \leq x \leq 2\\\left( x^2 - 3x + 2 \right)&, &x^2 - 3x + 2 \geq 0 \text{ or}, x \in \left( - \infty , 1 \right) \cup \left( 2, \infty \right)\end{cases}\]
\[ \therefore I = \int_0^2 \left( x^2 - 3x + 2 \right) d x\]
\[ \Rightarrow I = \int_0^1 \left( x^2 - 3x + 2 \right) d x - \int_1^2 \left( x^2 - 3x + 2 \right) d x\]
\[ \Rightarrow I = \left[ \frac{x^3}{3} - \frac{3 x^2}{2} + 2x \right]_0^1 - \left[ \frac{x^3}{3} - \frac{3 x^2}{2} + 2x \right]_1^2 \]
\[ \Rightarrow I = \frac{1}{3} - \frac{3}{2} + 2 - \left[ \frac{8}{3} - 6 + 4 - \frac{1}{3} + \frac{3}{2} - 2 \right]\]
\[ \Rightarrow I = \frac{1}{3} - \frac{3}{2} + 2 - \frac{8}{3} + 6 - 2 + \frac{1}{3} - \frac{3}{2}\]
\[ \Rightarrow I = 1\]