The area (in sq. units) of the region describ | vedclass

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(A) The region is bounded by the parabola $y = x^2 - 5x + 4$,the line $y = 1 - x$,and the line $y = 0$ (x-axis).First,find the intersection points:For

Vedclass · Accepted Answer

$\frac{19}{6}$

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(A) The region is bounded by the parabola $y = x^2 - 5x + 4$,the line $y = 1 - x$,and the line $y = 0$ (x-axis).First,find the intersection points:For $y = x^2 - 5x + 4$ and $y = 1 - x$:$x^2 - 5x + 4 = 1 - x \implies x^2 - 4x + 3 = 0 \implies (x-1)(x-3) = 0$.So,the intersection points are at $x=1$ and $x=3$. At $x=3$,$y = 1-3 = -2$.The region consists of two parts:$A_1$: Triangle bounded by vertices $(1,0), (3,0), (3,-2)$. Area $A_1 = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes (3-1) 	imes |-2| = \frac{1}{2} 	imes 2 	imes 2 = 2$.$A_2$: Area between the parabola and the x-axis from $x=3$ to $x=4$. Area $A_2 = |\int_{3}^{4} (x^2 - 5x + 4) dx| = |[\frac{x^3}{3} - \frac{5x^2}{2} + 4x]_3^4| = |(\frac{64}{3} - 40 + 16) - (9 - \frac{45}{2} + 12)| = |(\frac{64}{3} - 24) - (21 - 22.5)| = |-\frac{8}{3} - (-1.5)| = |-\frac{8}{3} + \frac{3}{2}| = |-\frac{16}{6} + \frac{9}{6}| = |-\frac{7}{6}| = \frac{7}{6}$.Total Area $= A_1 + A_2 = 2 + \frac{7}{6} = \frac{19}{6}$ sq. units.

The area (in sq. units) of the region described by $A = \{ (x,y) | y \ge x^2 - 5x + 4, x + y \ge 1, y \le 0 \}$ is:

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