The area bounded by the curve y=x^3-3x^2+2x a | vedclass

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(A) The given curve is $y = x^3 - 3x^2 + 2x$. To find the points where the curve intersects the $X$-axis, we set $y = 0$: $x(x^2 - 3x + 2) = 0$ $x(x -

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$\frac{1}{2}$

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(A) The given curve is $y = x^3 - 3x^2 + 2x$. To find the points where the curve intersects the $X$-axis, we set $y = 0$: $x(x^2 - 3x + 2) = 0$ $x(x - 1)(x - 2) = 0$ So, the curve intersects the $X$-axis at $x = 0, 1, 2$. The area is given by $\int_0^2 |y| dx = \int_0^1 (x^3 - 3x^2 + 2x) dx + \left| \int_1^2 (x^3 - 3x^2 + 2x) dx \right|$. First part: $\int_0^1 (x^3 - 3x^2 + 2x) dx = \left[ \frac{x^4}{4} - x^3 + x^2 \right]_0^1 = (\frac{1}{4} - 1 + 1) - 0 = \frac{1}{4}$. Second part: $\int_1^2 (x^3 - 3x^2 + 2x) dx = \left[ \frac{x^4}{4} - x^3 + x^2 \right]_1^2 = (\frac{16}{4} - 8 + 4) - (\frac{1}{4} - 1 + 1) = (4 - 8 + 4) - \frac{1}{4} = 0 - \frac{1}{4} = -\frac{1}{4}$. The absolute value is $|-\frac{1}{4}| = \frac{1}{4}$. Total area = $\frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$ square units.

The area bounded by the curve $y=x^3-3x^2+2x$ and the $X$-axis is (in square units)

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