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

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(A) We need to evaluate the integral $I = \int_0^3 |x^2 - 3x + 2| dx$. First, factor the quadratic expression: $x^2 - 3x + 2 = (x - 1)(x - 2)$. The ex

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$\frac{11}{6}$

Vedclass · Answer

(A) We need to evaluate the integral $I = \int_0^3 |x^2 - 3x + 2| dx$. First, factor the quadratic expression: $x^2 - 3x + 2 = (x - 1)(x - 2)$. The expression $(x - 1)(x - 2)$ is non-negative for $x \in [0, 1] \cup [2, 3]$ and negative for $x \in (1, 2)$. Thus, the integral can be split as: $I = \int_0^1 (x^2 - 3x + 2) dx - \int_1^2 (x^2 - 3x + 2) dx + \int_2^3 (x^2 - 3x + 2) dx$. Evaluating each integral: $\int (x^2 - 3x + 2) dx = \frac{x^3}{3} - \frac{3x^2}{2} + 2x$. For the first part: $[\frac{x^3}{3} - \frac{3x^2}{2} + 2x]_0^1 = (\frac{1}{3} - \frac{3}{2} + 2) - 0 = \frac{2 - 9 + 12}{6} = \frac{5}{6}$. For the second part: $-[\frac{x^3}{3} - \frac{3x^2}{2} + 2x]_1^2 = -[(\frac{8}{3} - 6 + 4) - (\frac{1}{3} - \frac{3}{2} + 2)] = -[(\frac{2}{3}) - (\frac{5}{6})] = -[\frac{4-5}{6}] = \frac{1}{6}$. For the third part: $[\frac{x^3}{3} - \frac{3x^2}{2} + 2x]_2^3 = (\frac{27}{3} - \frac{27}{2} + 6) - (\frac{8}{3} - 6 + 4) = (9 - 13.5 + 6) - (\frac{2}{3}) = 1.5 - \frac{2}{3} = \frac{3}{2} - \frac{2}{3} = \frac{9-4}{6} = \frac{5}{6}$. Adding these values: $I = \frac{5}{6} + \frac{1}{6} + \frac{5}{6} = \frac{11}{6}$.

$t \ (\text{in second})$	$0$	$2$	$4$	$6$	$8$	$10$
$v \ (\text{in m/s})$	$0$	$12$	$16$	$20$	$35$	$60$

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

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