Question

The area under the curve $y=x^2–3x+2$ with boundaries as x-axis and the ordinates x = 0, x = 3 is

• A. $\frac{3}{8}$
• B. $\frac{2}{3}$
• C. $\frac{3}{5}$
• D. $\frac{9}{2}$

Answer 1

The correct option is D $\frac{9}{2}$


$y=0\Rightarrow x^2-3x+2=0\Rightarrow (x-1)(x-2)=0\Rightarrow x=1,2$
Required area
$={\int }_0^1(x^2-3x+2)dx-{\int }_1^2(x^2-3x+2)dx+{\int }_2^3(x^2-3x+2)dx$
${[\frac{x^3}{3}-\frac{3x^2}{2}+2x]}_0^1-{[\frac{x^3}{3}-\frac{3x^2}{2}+2x]}_1^2+{[\frac{x^3}{3}-\frac{3x^2}{2}+2x]}_2^3$
$=[\frac{1}{3}-\frac{3}{2}+2]-[\frac{8}{3}-6+4]+[\frac{1}{3}-\frac{3}{2}+2][9-\frac{27}{2}-+6]-[\frac{8}{3}-6+4]$
$=2\left[\frac{2-9+12}{6}\right]+2\left[\frac{8-6}{3}\right]+\left[\frac{30-27}{2}\right]=\frac{10}{6}+\frac{4}{3}+\frac{3}{2}=\frac{10+8+9}{6}=\frac{27}{6}=\frac{9}{2}$sq. unit