question_answer

Find the area of the region \[\{(x,\,\,y):0\le y\le {{x}^{2}}+1,\,\,0\le y\le x+1,\,\,0\le x\le 2\}.\]  

Answer:

Given equations of curves and lines are \[y={{x}^{2}}+1\]                 ?(i) \[y=x+1\]                   ?(ii) and                   \[x=2\]                       ?(iii) Eq. (i) represents a parabola with vertex at A (0, 1) and axis along the positive direction of Y-axis. Eq. (ii) represents a line which intersects the coordinate axes and Eq. (iii) represents a straight line parallel to Y-axis. On solving Eqs. (i) and (ii), we get \[{{x}^{2}}+1=x+1\] \[\Rightarrow \]\[{{x}^{2}}-x=0\Rightarrow x\,(x-1)=0\Rightarrow x=1\]and \[x=0\] \[\therefore \]When \[x=0,\]then \[y=0+1=1\][using Eq. (ii)] When \[x=1,\]then\[y=1+1=2\]  [using Eq. (ii)] So, points of intersection of Eqs. (i) and (ii) are A (0, 1)and B (1, 2). As \[0\le y\le {{x}^{2}}+1,\]which represents the area below the parabola and above X-axis. \[y=x+1\]represents a line AB. \[0\le y\le x+1\]represents the region below the line AB and above X-axis. Also, \[0\le x\le 2\]represents the area between the parallel lines \[x=0\] and\[x=2\]. \[\therefore \]The region common to \[\left. \begin{matrix}    0\le y\le {{x}^{2}}+1  \\    0\le y\le x+1  \\    0\le x\le 2  \\ \end{matrix} \right\}\]is shown by shaded region \[\therefore \]Required area = Area of the region OACBFO \[+\] Area of region BDEFB \[=\int_{0}^{1}{\,\,\,{{y}_{(parabpla)}}dx+\int_{1}^{2}{\,\,\,\,{{y}_{\,\,\,\,(line)}}dx}}\] \[=\int_{0}^{1}{({{x}^{2}}+1)\,dx+\int_{1}^{2}{(x+1)\,dx}}\] \[=\left[ \frac{{{x}^{3}}}{3}+x \right]_{0}^{1}+\left[ \frac{{{x}^{2}}}{2}+x \right]_{1}^{2}\] \[=\left[ \left( \frac{1}{3}+1 \right)-0 \right]+\left[ \left( \frac{4}{2}+2 \right)-\left( \frac{1}{2}+1 \right) \right]\] \[=\frac{4}{3}+4-\frac{3}{2}=\frac{4}{3}+\frac{5}{2}=\frac{8+15}{6}=\frac{23}{6}\]sq units