Question

Area of region enclosed by curve $y=|x-3|$. X-axis and lines $x=0,x=2$ is _________.

• A. $\frac{3}{2}$
• B. $4$
• C. $\frac{9}{2}$
• D. $2$

Answer 1

Answer: (B)

$4$

We take $y=0$ to get point at intersection of the curve $y=|x-3|$ with X-axis.
$\therefore x-3=0$
$\therefore x=3$
$\therefore$ Curve intersects X-axis at $(3,0)$
Required area $=\stackrel{2}{\underset{0}{\int }}|x-3|dx$
$=\left|\stackrel{2}{\underset{0}{\int }}\right(x-3\left)dx\right|$
Here $x\in |0,2|$ so we have $x-3<0$
$=\stackrel{2}{\underset{0}{\int }}(3-x)dx$
$={(3x-\frac{x^2}{2})}_0^2$
$=(6-\frac{4}{2})-0$
$=\frac{12-4}{2}$
$=\frac{8}{2}$
$\therefore$ Required area $=4$.