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. $\frac{3}{5}$ sq. units
• B. $2$ sq. units
• C. $\frac{11}{6}$ sq. units
• D. $\frac{4}{5}$ sq. units

Answer 1

The correct option is C $\frac{11}{6}$ sq. units

Given curve is $x^2-3x+2=0$

The roots of this quadratic equation are $1$ and $2$

Thus, the area under the curve and X-axis and the ordinates $x=0$ $x=3$ is

$A={\int }_0^3(x^2-3x+2)$

$A={\int }_0^1(x^2-3x+2)-{\int }_1^2(x^2-3x+2)+{\int }_2^3(x^2-3x+2)$ ....... (Between $x=1$ to $x=2$, the curve is below X-axis)

After integrating and substituting upper and lower limits, we get

$A=\frac{5}{6}+\frac{1}{6}+\frac{5}{6}$

$\therefore A=\frac{11}{6}$