int_0^3 |x^2 - 3x + 2| dx = | vedclass

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Question

(C) Let $I = \int_0^3 |x^2 - 3x + 2| dx$. Since $x^2 - 3x + 2 = (x - 1)(x - 2)$,the expression inside the modulus changes sign at $x = 1$ and $x = 2$.

Vedclass · Accepted Answer

$11/6$

Vedclass · Answer

(C) Let $I = \int_0^3 |x^2 - 3x + 2| dx$. Since $x^2 - 3x + 2 = (x - 1)(x - 2)$,the expression inside the modulus changes sign at $x = 1$ and $x = 2$. For $x \in [0, 1]$,$x^2 - 3x + 2 \ge 0$. For $x \in (1, 2)$,$x^2 - 3x + 2 For $x \in [2, 3]$,$x^2 - 3x + 2 \ge 0$. Thus,$I = \int_0^1 (x^2 - 3x + 2) dx - \int_1^2 (x^2 - 3x + 2) dx + \int_2^3 (x^2 - 3x + 2) dx$. Evaluating the integral $\int (x^2 - 3x + 2) dx = \frac{x^3}{3} - \frac{3x^2}{2} + 2x + C$. $I = [\frac{x^3}{3} - \frac{3x^2}{2} + 2x]_0^1 - [\frac{x^3}{3} - \frac{3x^2}{2} + 2x]_1^2 + [\frac{x^3}{3} - \frac{3x^2}{2} + 2x]_2^3$. $I = (\frac{1}{3} - \frac{3}{2} + 2) - [(\frac{8}{3} - 6 + 4) - (\frac{1}{3} - \frac{3}{2} + 2)] + [(\frac{27}{3} - \frac{27}{2} + 6) - (\frac{8}{3} - 6 + 4)]$. $I = (\frac{2 - 9 + 12}{6}) - [(\frac{2}{3}) - (\frac{5}{6})] + [(\frac{3}{2}) - (\frac{2}{3})]$. $I = \frac{5}{6} - [-\frac{1}{6}] + \frac{5}{6} = \frac{5}{6} + \frac{1}{6} + \frac{5}{6} = \frac{11}{6}$.

$\int_0^3 |x^2 - 3x + 2| dx = $

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