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  • Using integration, find the area of the region bounded by the triangle whose vertices are (2, – 2), (4, 5) and (6, 2).  

Using integration, find the area of the region bounded by the triangle whose vertices are (2, – 2), (4, 5) and (6, 2).

 

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Area under triangle ABC

Equation of line AB 

\\ y+2 = \frac{5+2}{4-2}(x-2) \\\\ 2y+4 = 7x-14 \\\\ \mathrm{x}=\frac{2}{7}(\mathrm{y}+9) \\

Equation of line BC 

\\ y-5 = \frac{2-5}{6-4}(x-4) \\\\ 2y-10 = -3x+12 \\\\ \mathrm{x}=-\frac{2}{3}(\mathrm{y}-11) \\

Equation of line AC

\\ y+2 = \frac{2+2}{6-2}(x-2) \\\\ 4y+8 = 4x-8 \\\\ \mathrm{x}=\mathrm{y}+4 \\

Area of triangle ABC

\\=\int_{-2}^{2}(\mathrm{y}+4) \mathrm{d} \mathrm{y}+\left(\frac{-2}{3}\right) \int_{2}^{5}(\mathrm{y}-11) \mathrm{d} \mathrm{y}-\int_{-2}^{5} \frac{2}{7}(\mathrm{y}+9) \mathrm{dy} \\\\ =\frac{1}{2}\left[(\mathrm{y}+4)^{2}\right]_{-2}^{2}-\frac{1}{3}\left[(\mathrm{y}-11)^{2}\right]_{2}^{5}-\frac{1}{7}\left[(\mathrm{y}+9)^{2}\right]_{-2}^{5} \\\\ =\frac{1}{2}(36-4)-\frac{1}{3}(36-81) - \frac{1}{7}(196-49) \\\\ =16+15-21\\=10

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Safeer PP

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