91Ó°ÊÓ

The given equation is \( y = (1-x)(x-2) \). This is a quadratic equation in the form \( y = ax^2 + bx + c \) when expanded. Let's expand it to understand its nature.

Expand the equation \( y = (1-x)(x-2) \): \[ y = (1-x)(x-2) = 1 \cdot x - 2 - x^2 + 2x = -x^2 + 3x - 2 \]The equation is \( y = -x^2 + 3x - 2 \). This is a downward-opening parabola.

Set \( y = 0 \) to find the x-intercepts of the parabola:\[-x^2 + 3x - 2 = 0 \]Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = -1, b = 3, c = -2 \).\[ x = \frac{-3 \pm \sqrt{3^2 - 4(-1)(-2)}}{2(-1)} \]\[ x = \frac{-3 \pm \sqrt{9 - 8}}{-2} \]\[ x = \frac{-3 \pm 1}{-2} \]The roots are \( x = 1 \) and \( x = 2 \).

The area we seek is between the roots \( x = 1 \) and \( x = 2 \). Compute the definite integral of \( -x^2 + 3x - 2 \) from \( x = 1 \) to \( x = 2 \):\[ \int_{1}^{2} (-x^2 + 3x - 2) \, dx \]Evaluate the integral:\[ = \left[ -\frac{x^3}{3} + \frac{3x^2}{2} - 2x \right]_{1}^{2} \]\[ = \left( -\frac{2^3}{3} + \frac{3 \cdot 2^2}{2} - 2 \cdot 2 \right) - \left( -\frac{1^3}{3} + \frac{3 \cdot 1^2}{2} - 2 \cdot 1 \right) \]\[ = \left( -\frac{8}{3} + 6 - 4 \right) - \left( -\frac{1}{3} + \frac{3}{2} - 2 \right) \]\[ = \left(-\frac{8}{3} + 2\right) - \left(-\frac{1}{3} + \frac{3}{2} - 2\right) \]\[ = -\frac{8}{3} + \frac{6}{3} + \frac{1}{3} - \frac{3}{2} + 2 \]\[ = \frac{-8 + 6 + 1}{3} + \frac{-3}{2} + 2 \]\[ = \frac{-1}{3} + \frac{4}{2} - \frac{3}{2} \]\[ = \frac{-1}{3} + \frac{1}{2} \]Convert \(\frac{1}{2}\) to a common denominator:\[ = \frac{-2}{6} + \frac{3}{6} = \frac{1}{6} \]The area above the x-axis and below the curve is \( \frac{1}{6} \).

Draw a graph with the x-axis and the parabola \( y = -x^2 + 3x - 2 \). Label the intersections at \( x = 1 \) and \( x = 2 \) on the x-axis. Shade the area between these points and above the x-axis.