Class 12 Applications Of Integrals · Very Long Answer5 Marksarea_by_integration

Class 12 Maths · Applications Of Integrals · Very Long Answer #23

Using integration, find the area bounded by 3x−2y+1=0, 2x+3y−21=0 and x−5y+9=0.

✅ Step-by-Step Solution

Final Answer: \(\frac{13}{2}\) square units

In increasing x order, the vertices are (1,2), (3,5), (6,3). The left edge, right edge and long edge have equations y=\(\frac{3 x}{2} + \frac{1}{2}\), y=\(7 - \frac{2 x}{3}\), y=\(\frac{x}{5} + \frac{9}{5}\), respectively.
Use upper minus lower strip widths: \(\frac{13 x}{10} - \frac{13}{10}\) for x∈\(\left[1, 3\right]\); \(\frac{26}{5} - \frac{13 x}{15}\) for x∈\(\left[3, 6\right]\).
Hence A=\(\int_{1}^{3}(\frac{13 x}{10} - \frac{13}{10})\,dx\) + \(\int_{3}^{6}(\frac{26}{5} - \frac{13 x}{15})\,dx\).
Primitive \(\frac{13 x^{2}}{20} - \frac{13 x}{10}\), endpoint difference \(\frac{13}{5}\); Primitive \(- \frac{13 x^{2}}{30} + \frac{26 x}{5}\), endpoint difference \(\frac{39}{10}\).
Therefore the area is \(\frac{13}{2}\) square units.

[1 mark] In increasing x order, the vertices are (1,2), (3,5), (6,3). The left edge, right edge and long edge have equations y=\(\frac{3 x}{2} + \frac{1}{2}\), y=\(7 - \frac{2 x}{3}\), y=\(\frac{x}{5} + \frac{9}{5}\), respectively.
[1 mark] Use upper minus lower strip widths: \(\frac{13 x}{10} - \frac{13}{10}\) for x∈\(\left[1, 3\right]\); \(\frac{26}{5} - \frac{13 x}{15}\) for x∈\(\left[3, 6\right]\).
[1 mark] Hence A=\(\int_{1}^{3}(\frac{13 x}{10} - \frac{13}{10})\,dx\) + \(\int_{3}^{6}(\frac{26}{5} - \frac{13 x}{15})\,dx\).
[1 mark] Primitive \(\frac{13 x^{2}}{20} - \frac{13 x}{10}\), endpoint difference \(\frac{13}{5}\); Primitive \(- \frac{13 x^{2}}{30} + \frac{26 x}{5}\), endpoint difference \(\frac{39}{10}\).
[1 mark] Therefore the area is \(\frac{13}{2}\) square units.

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