The value of 12 int limits_0^3 left| x^2 - 3x | vedclass

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(C) Let $I = 12 \int_0^3 |x^2 - 3x + 2| dx$. First,factor the quadratic expression: $x^2 - 3x + 2 = (x - 1)(x - 2)$. The expression $(x - 1)(x - 2)$ i

Vedclass · Accepted Answer

$22$

Vedclass · Answer

(C) Let $I = 12 \int_0^3 |x^2 - 3x + 2| dx$. First,factor the quadratic expression: $x^2 - 3x + 2 = (x - 1)(x - 2)$. The expression $(x - 1)(x - 2)$ is positive on $[0, 1)$,negative on $(1, 2)$,and positive on $(2, 3]$. Thus,we split the integral: $I = 12 \left[ \int_0^1 (x^2 - 3x + 2) dx + \int_1^2 -(x^2 - 3x + 2) dx + \int_2^3 (x^2 - 3x + 2) dx \right]$. Evaluating each integral: $\int (x^2 - 3x + 2) dx = \frac{x^3}{3} - \frac{3x^2}{2} + 2x$. For $[0, 1]$: $[\frac{1}{3} - \frac{3}{2} + 2] - [0] = \frac{2 - 9 + 12}{6} = \frac{5}{6}$. For $[1, 2]$: $-[(\frac{8}{3} - 6 + 4) - (\frac{1}{3} - \frac{3}{2} + 2)] = -[(\frac{2}{3}) - (\frac{5}{6})] = -[\frac{4-5}{6}] = \frac{1}{6}$. For $[2, 3]$: $[(9 - \frac{27}{2} + 6) - (\frac{8}{3} - 6 + 4)] = [15 - 13.5 - \frac{2}{3}] = [1.5 - \frac{2}{3}] = \frac{3}{2} - \frac{2}{3} = \frac{9-4}{6} = \frac{5}{6}$. Summing these: $I = 12 \left( \frac{5}{6} + \frac{1}{6} + \frac{5}{6} \right) = 12 \left( \frac{11}{6} \right) = 22$.

The value of $12 \int \limits_0^3 \left| x^2 - 3x + 2 \right| dx$ is $.............$

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