91Ó°ÊÓ

The x-axis intersects the curve when \(y = 0\). So, we need to solve the equation \(x(x-1)(x-2) = 0\) for \(x\). This equation has three solutions, which are the x-intercepts: \(x = 0\), \(x = 1\), and \(x = 2\).

To find the area bounded by the curve and the x-axis, we will integrate the function \(y(x) = x(x-1)(x-2)\) with respect to \(x\). Since the curve is above the x-axis between the points \(x = 0\) and \(x = 1\), and below the x-axis between the points \(x = 1\) and \(x = 2\), we need to split the integral into two parts and take the absolute value of each part. So, the area is given by: $$ \text{Area} = \int_{0}^{1} |x(x-1)(x-2)| \, dx + \int_{1}^{2} |x(x-1)(x-2)| \, dx $$

To evaluate the integral, we can first find the expression for each part of the integral. For \(0\le x\le1\), \(x(x-1)(x-2)\) is non-negative (as x is between 0 and 1). So, the absolute value is not needed in this case: $$ \int_{0}^{1} x(x-1)(x-2) \, dx $$ For \(1\le x\le2\), \(x(x-1)(x-2)\) is non-positive (as x is between 1 and 2). So, we need to multiply the expression by -1 to get the absolute value: $$ \int_{1}^{2} -(x(x-1)(x-2)) \, dx $$ Now, we can evaluate both integrals. We know that, $$ \int x(x-1)(x-2) \, dx = \frac{1}{4}x^4 - \frac{5}{3}x^3 + \frac{9}{2}x^2 - \frac{11}{12}x + C $$ So, $$ \text{Area} = \left(\frac{1}{4}(1)^4 - \frac{5}{3}(1)^3 + \frac{9}{2}(1)^2 - \frac{11}{12}(1)\right) - \left(\frac{1}{4}(0)^4 - \frac{5}{3}(0)^3 + \frac{9}{2}(0)^2 - \frac{11}{12}(0)\right) \\ + \left(\frac{1}{4}(2)^4 - \frac{5}{3}(2)^3 + \frac{9}{2}(2)^2 - \frac{11}{12}(2)\right) - \left(\frac{1}{4}(1)^4 - \frac{5}{3}(1)^3 + \frac{9}{2}(1)^2 - \frac{11}{12}(1)\right) $$ After evaluating the expression, we get: $$ \text{Area} = \frac{9}{4} $$ So, the area bounded by the curve \(y = x(x-1)(x-2)\) and the x-axis is \(\frac{9}{4}\).