Question

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

• A. $\frac{17}{6}$
• B. $\frac{13}{6}$
• C. $\frac{19}{6}$
• D. $\frac{7}{2}$

Answer 1

Answer: (C)

$\frac{19}{6}$

$y\ge x^2-5x+4=(x-\frac{5}{2}{)}^2-\frac{9}{4}$
$y+\frac{9}{4}\ge (x-\frac{5}{2}{)}^2$
This is a parabola with vertex at $(\frac{5}{2},-\frac{9}{4})$
And other two curves are
$x+y\ge 1$ and $y\le 0$
Draw it. And From the figure we can see
Area of the shaded region AED= Area of triangle ADE+ area of the region EBD
Area of triangle ABD$=\frac{1}{2}\times 2\times 2=2$

Area of region EBD $\left|{\int }_3^{4}ydx\right|=\left|{\int }_3^4\right(x^2-5x+4\left)dx\right|$

$=\left|{[\frac{x^3}{3}-\frac{5x^2}{2}+4x]}_3^4\right|$

$=|\frac{263}{3}-45|=\frac{7}{6}$
Hence, Total Area $=2+\frac{7}{6}=\frac{19}{6}$