int_{1}^{3} (x - 1)(x - 2)(x - 3) , dx = | vedclass

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(D) Let $I = \int_{1}^{3} (x - 1)(x - 2)(x - 3) \, dx$. Expanding the integrand: $(x - 1)(x - 2)(x - 3) = (x^2 - 3x + 2)(x - 3) = x^3 - 3x^2 - 3x^2 +

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(D) Let $I = \int_{1}^{3} (x - 1)(x - 2)(x - 3) \, dx$. Expanding the integrand: $(x - 1)(x - 2)(x - 3) = (x^2 - 3x + 2)(x - 3) = x^3 - 3x^2 - 3x^2 + 9x + 2x - 6 = x^3 - 6x^2 + 11x - 6$. Now,integrate term by term: $I = \left[ \frac{x^4}{4} - \frac{6x^3}{3} + \frac{11x^2}{2} - 6x \right]_{1}^{3} = \left[ \frac{x^4}{4} - 2x^3 + \frac{11x^2}{2} - 6x \right]_{1}^{3}$. Evaluating at the upper limit $x = 3$: $\left( \frac{81}{4} - 2(27) + \frac{11(9)}{2} - 6(3) \right) = \frac{81}{4} - 54 + \frac{99}{2} - 18 = \frac{81 + 198}{4} - 72 = \frac{279}{4} - 72 = \frac{279 - 288}{4} = -\frac{9}{4}$. Evaluating at the lower limit $x = 1$: $\left( \frac{1}{4} - 2(1) + \frac{11}{2} - 6(1) \right) = \frac{1}{4} - 2 + \frac{11}{2} - 6 = \frac{1 + 22}{4} - 8 = \frac{23}{4} - 8 = \frac{23 - 32}{4} = -\frac{9}{4}$. Therefore,$I = -\frac{9}{4} - (-\frac{9}{4}) = 0$.

$\int_{1}^{3} (x - 1)(x - 2)(x - 3) \, dx = $

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