The area enclosed by the curve y = (x - 1)(x | vedclass

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(C) The curve is $y = (x - 1)(x - 2)(x - 3) = x^3 - 6x^2 + 11x - 6$.The area is bounded by the $y$-axis $(x=0)$,the $x$-axis $(y=0)$,and the line $x=3

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$\frac{11}{4}$

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(C) The curve is $y = (x - 1)(x - 2)(x - 3) = x^3 - 6x^2 + 11x - 6$.The area is bounded by the $y$-axis $(x=0)$,the $x$-axis $(y=0)$,and the line $x=3$.The roots of the curve are $x=1, 2, 3$. The curve intersects the $x$-axis at these points.Area $A = \int_{0}^{3} |y| dx = \int_{0}^{1} |x^3 - 6x^2 + 11x - 6| dx + \int_{1}^{2} |x^3 - 6x^2 + 11x - 6| dx + \int_{2}^{3} |x^3 - 6x^2 + 11x - 6| dx$.Let $I(x) = \int (x^3 - 6x^2 + 11x - 6) dx = \frac{x^4}{4} - 2x^3 + \frac{11x^2}{2} - 6x$.$1$. For $x \in [0, 1]$,$y $2$. For $x \in [1, 2]$,$y > 0$,so Area $A_2 = I(2) - I(1) = [(\frac{16}{4} - 2(8) + \frac{11(4)}{2} - 6(2)) - (-\frac{9}{4})] = [(4 - 16 + 22 - 12) + \frac{9}{4}] = [-2 + \frac{9}{4}] = \frac{1}{4}$.$3$. For $x \in [2, 3]$,$y Total Area $= A_1 + A_2 + A_3 = \frac{9}{4} + \frac{1}{4} + \frac{1}{4} = \frac{11}{4}$.

The area enclosed by the curve $y = (x - 1)(x - 2)(x - 3)$ between the coordinate axes and the ordinate at $x = 3$ is:

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