question_answer

Using integration, find the area of the triangular region whose vertices are

P (1, 0), Q (2, 2) and R (3, 1).

OR

Using integration find the area of the following region.

\[\{(x,\,\,y):|x-1|\,\,\le y\le \sqrt{5-{{x}^{2}}}\}\]

Answer:

Given vertices of triangular region are \[P\,(1,\,\,0),\]\[Q\,(2,\,\,2)\]and\[R\,(3,\,\,1)\]. \[\therefore \]Equation of the line joining points \[({{x}_{1}},\,\,{{y}_{1}})\]and \[({{x}_{2}},\,\,{{y}_{2}})\]is \[(y-{{y}_{1}})=\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}+{{x}_{1}}}(x-{{x}_{1}})\] Equation of line PQ is \[y-0=\frac{2-0}{2-1}(x-1)\] \[\Rightarrow \]               \[y=2\,(x-1)\] Equation of line QR is \[y-2=\frac{1-2}{3-2}(x-2)\]\[\Rightarrow y=2-(x-2)=4-x\] Equation of line PR is \[y-0=\frac{1-0}{3-1}(x-1)\Rightarrow y=0+\frac{1}{2}(x-1)=\frac{1}{2}(x-1)\] Area of \[\Delta \,PQR=Area\,\,(PMQ)+Area\,\,(QMNR)-Area\,(PNR)\]\[=\int_{1}^{2}{2\,(x-1)\,\,dx+\int_{2}^{3}{(4-x)\,\,dx-\int_{1}^{3}{\frac{1}{2}(x-1)}}}\] \[=2\left[ \frac{{{x}^{2}}}{2}-x \right]_{1}^{2}+\left[ 4x-\frac{{{x}^{2}}}{2} \right]_{2}^{3}-\frac{1}{2}\left[ \frac{{{x}^{2}}}{2}-x \right]_{1}^{3}\] \[=\left[ \frac{x}{2}\sqrt{5-{{x}^{2}}}+\frac{5}{2}{{\sin }^{-\,1}}\frac{x}{\sqrt{5}} \right]_{-\,1}^{2}-\left[ -\frac{{{x}^{2}}}{2}+x \right]_{-\,1}^{1}-\left[ \frac{{{x}^{2}}}{2}-x \right]_{1}^{2}\] \[-\frac{1}{2}\left[ \left( \frac{9}{2}-3 \right)-\left( \frac{1}{2}-1 \right) \right]\] \[=1+\frac{3}{2}-1=\frac{3}{2}sq\,\,units\] Corresponding equations are\[y=\,\,|x-1|\]and \[y=\sqrt{5-{{x}^{2}}}\] Eliminating y, we get\[|x-1|\,\,=\sqrt{5-{{x}^{2}}}\] \[\Rightarrow \]   \[{{x}^{2}}+1-2x=5-{{x}^{2}}\Rightarrow 2,\,\,-\,1\] \[\therefore \]Area\[=\int_{-\,1}^{2}{\sqrt{5-{{x}^{2}}}dx}-\int_{-\,1}^{1}{-\,(x-1)\,dx-\int_{1}^{2}{(x-1)}}\,dx\] \[=\left[ \frac{x}{2}\sqrt{5-{{x}^{2}}}+\frac{5}{2}{{\sin }^{-\,1}}\frac{x}{\sqrt{5}} \right]_{-\,1}^{2}\] \[-\left[ -\frac{{{x}^{2}}}{2}+x \right]_{-\,1}^{1}-\left[ \frac{{{x}^{2}}}{2}-x \right]_{1}^{2}\] \[=\left( 1+\frac{5}{2}{{\sin }^{-\,1}}\frac{2}{\sqrt{5}} \right)-\left[ -\,1+\frac{5}{2}{{\sin }^{-\,1}}\left( -\frac{1}{\sqrt{5}} \right) \right]\] \[-\left[ -\frac{1}{2}+1 \right]\] \[+\left[ -\frac{1}{2}-1 \right]-[2-2]+\left( \frac{1}{2}-1 \right)\] \[=1+\frac{5}{2}{{\sin }^{-\,1}}\frac{2}{\sqrt{5}}+1+\frac{5}{2}{{\sin }^{-\,1}}\frac{1}{\sqrt{5}}-\frac{1}{2}-\frac{3}{2}-\frac{1}{2}\] \[=\frac{5}{2}{{\sin }^{-\,1}}\left[ \frac{2}{\sqrt{5}}\sqrt{1-\frac{1}{5}}+\frac{1}{\sqrt{5}}\sqrt{1-\frac{4}{5}} \right]-\frac{1}{2}\] \[=\frac{5}{2}{{\sin }^{-\,1}}\left[ \frac{4}{5}+\frac{1}{5} \right]-\frac{1}{2}=\frac{5}{2}\cdot \frac{\pi }{2}-\frac{1}{2}=\frac{1}{4}(5\pi -2)\]sq units