Question

Let $f\left(x\right)=x^2-5x+6$ be a function $\forall x\in R$.

Area bounded by $f\left(x\right)$ between the abcissae $x=2$ & $x=3$ lying below $x$-axis is

• A. $\frac{1}{6}$ square unit
• B. $\frac{2}{3}$ square unit
• C. $\frac{1}{3}$ square unit
• D. $\frac{5}{6}$ square unit

Answer 1

Answer: (A)

$\frac{1}{6}$ square unit

The area lies below the $x-$axis between the abcissae $x=2$ and $x=3$.
$\therefore$ Required area $=\left|{\int }_2^3(x^2-5x+6)dx\right|$
$=\left|{(\frac{x^3}{3}-\frac{5x^2}{2}+6x)}_2^3\right|=|(\frac{27}{3}-\frac{45}{2}+18)-(\frac{8}{3}-10+12)|$
$=|(27-\frac{45}{2})-\left(\frac{14}{3}\right)|=|\frac{9}{2}-\frac{14}{3}|=\left|\frac{27-28}{6}\right|$
$=\frac{1}{6}$ square unit