Question

Area of the region bounded by $y=x^2-3x+2$, then $x-$axis and the ordinates $x=0,x=3$ is

• A. $\frac{11}{6}$
• B. $\frac{3}{2}$
• C. $2$
• D. $None$

Answer 1

The correct option is B $\frac{3}{2}$

$y=x^2-3x+2$

$y-2=x^2-3x+\frac{9}{4}-\frac{9}{4}$

$(x-\frac{3}{2}{)}^2=(y+\frac{1}{4})$

Area bounded $={\int }_0^{\frac{7}{4}}ydx+\left|{\int }_{\frac{7}{4}}^{\frac{5}{4}}ydx\right|+{\int }_{\frac{5}{4}}^{3}ydx$

$={\int }_0^{\frac{7}{4}}\left(x^2-3x+2\right)dx+\left|{\int }_{\frac{7}{4}}^{\frac{5}{4}}\left(x^2-3x+2\right)dx\right|+{\int }_{\frac{5}{4}}^3\left(x^2-3x+2\right)dx$

$={\left[\frac{x^3}{3}-\frac{3x^2}{2}+2x\right]}_0^{\frac{7}{4}}+\left|{\left[\frac{x^3}{3}-\frac{3x^2}{2}+2x\right]}_{\frac{7}{4}}^{\frac{5}{4}}\right|+\left|{\left[\frac{x^3}{3}-\frac{3x^2}{2}+2x\right]}_{\frac{5}{4}}^3\right|$

$=\left(\frac{343}{64\times 3}-\frac{3\times 49}{2\times 16}+\frac{7}{2}\right)+\left|\left(\frac{125}{3\times 64}-\frac{3\times 25}{2\times 16}+\frac{5}{2}\right)-\left(\frac{343}{64\times 3}-\frac{3\times 49}{2\times 16}+\frac{7}{2}\right)\right|+\left(9-\frac{27}{2}+6\right)-\left(\frac{125}{3\times 64}-\frac{3\times 25}{2\times 16}+\frac{5}{2}\right)$

$=\frac{3}{2}$