Class 12 Applications Of Integrals · Very Long Answer5 Marksarea_by_integration

Class 12 Maths · Applications Of Integrals · Very Long Answer #44

Using integration, find the area of the triangle with vertices (2,3), (4,7), (6,2).

✅ Step-by-Step Solution

Final Answer: \(9\) square units

In increasing x order, the vertices are (2,3), (4,7), (6,2). The left edge, right edge and long edge have equations y=\(2 x - 1\), y=\(17 - \frac{5 x}{2}\), y=\(\frac{7}{2} - \frac{x}{4}\), respectively.
Use upper minus lower strip widths: \(\frac{9 x}{4} - \frac{9}{2}\) for x∈\(\left[2, 4\right]\); \(\frac{27}{2} - \frac{9 x}{4}\) for x∈\(\left[4, 6\right]\).
Hence A=\(\int_{2}^{4}(\frac{9 x}{4} - \frac{9}{2})\,dx\) + \(\int_{4}^{6}(\frac{27}{2} - \frac{9 x}{4})\,dx\).
Primitive \(\frac{9 x^{2}}{8} - \frac{9 x}{2}\), endpoint difference \(\frac{9}{2}\); Primitive \(- \frac{9 x^{2}}{8} + \frac{27 x}{2}\), endpoint difference \(\frac{9}{2}\).
Therefore the area is \(9\) square units.

[1 mark] In increasing x order, the vertices are (2,3), (4,7), (6,2). The left edge, right edge and long edge have equations y=\(2 x - 1\), y=\(17 - \frac{5 x}{2}\), y=\(\frac{7}{2} - \frac{x}{4}\), respectively.
[1 mark] Use upper minus lower strip widths: \(\frac{9 x}{4} - \frac{9}{2}\) for x∈\(\left[2, 4\right]\); \(\frac{27}{2} - \frac{9 x}{4}\) for x∈\(\left[4, 6\right]\).
[1 mark] Hence A=\(\int_{2}^{4}(\frac{9 x}{4} - \frac{9}{2})\,dx\) + \(\int_{4}^{6}(\frac{27}{2} - \frac{9 x}{4})\,dx\).
[1 mark] Primitive \(\frac{9 x^{2}}{8} - \frac{9 x}{2}\), endpoint difference \(\frac{9}{2}\); Primitive \(- \frac{9 x^{2}}{8} + \frac{27 x}{2}\), endpoint difference \(\frac{9}{2}\).
[1 mark] Therefore the area is \(9\) square units.

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