1 question with graph

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{y}={\mathrm{x}}^2-3\mathrm{x}+2=\left(\mathrm{x}-1\right)\left(\mathrm{x}-2\right) \\ \mathrm{Hence} \mathrm{roots} \mathrm{of} \mathrm{parabola} \mathrm{are} \mathrm{x}=1,2 \\ \mathrm{Coeffcient} \mathrm{of} {\mathrm{x}}^2>0, \mathrm{hence} \mathrm{opening} \mathrm{is} \mathrm{upward}. \end{gathered}$$


$$\begin{gathered} \left(\mathrm{i}\right) \mathrm{Area}=\left|{\int }_1^2 \left({\mathrm{x}}^2-3\mathrm{x}+2\right)\mathrm{dx}\right| \\ =\left|{\left[\frac{{\mathrm{x}}^3}{3}-\frac{3{\mathrm{x}}^2}{2}+2\mathrm{x}\right]}_1^2\right| \\ =\left|\left[\frac{8}{3}-\frac{12}{2}+4\right]-\left[\frac{1}{3}-\frac{3}{2}+2\right]\right| \\ =\left|\frac{7}{3}-\frac{9}{2}+2\right| \\ =\left|\frac{14-27+12}{6}\right| \\ =\left|-\frac{1}{6}\right| \\ =\frac{1}{6} \left(\mathrm{Answer}\right) \\ \left(\mathrm{ii}\right) \\ \mathrm{From} \mathrm{x}=0 \mathrm{to} \mathrm{x}=1 \mathrm{graph} \mathrm{is} \mathrm{above} \mathrm{x}-\mathrm{axis} \mathrm{and} \mathrm{x}=1 \mathrm{and} \mathrm{x}=2 \mathrm{graph} \mathrm{is} \mathrm{below} \\ \mathrm{x}-\mathrm{axis} \\ \mathrm{Area}={\int }_0^1 \left({\mathrm{x}}^2-3\mathrm{x}+2\right).\mathrm{dx}+\left|{\int }_1^2 \left({\mathrm{x}}^2-3\mathrm{x}+2\right)\mathrm{dx}\right| \\ ={\left[\frac{{\mathrm{x}}^3}{3}-\frac{3{\mathrm{x}}^2}{2}+2\mathrm{x}\right]}_0^1+\frac{1}{6} \\ =\left[\frac{1}{3}-\frac{3}{2}+2\right]-\left[0-0+0\right]+\frac{1}{6} \\ =\frac{2-9+12}{6}+\frac{1}{6} \\ =\frac{5}{6}+\frac{1}{6} \\ =1 \left(\mathrm{Answer}\right) \end{gathered}$$

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