Class 12 Applications Of Integrals · Very Long Answer5 Marksarea_by_integration

Class 12 Maths · Applications Of Integrals · Very Long Answer #50

Using integration, find the area of the triangle with vertices (2,-2), (4,3), (1,2).

✅ Step-by-Step Solution

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

In increasing x order, the vertices are (1,2), (2,-2), (4,3). The left edge, right edge and long edge have equations y=\(6 - 4 x\), y=\(\frac{5 x}{2} - 7\), y=\(\frac{x}{3} + \frac{5}{3}\), respectively.
Use upper minus lower strip widths: \(\frac{13 x}{3} - \frac{13}{3}\) for x∈\(\left[1, 2\right]\); \(\frac{26}{3} - \frac{13 x}{6}\) for x∈\(\left[2, 4\right]\).
Hence A=\(\int_{1}^{2}(\frac{13 x}{3} - \frac{13}{3})\,dx\) + \(\int_{2}^{4}(\frac{26}{3} - \frac{13 x}{6})\,dx\).
Primitive \(\frac{13 x^{2}}{6} - \frac{13 x}{3}\), endpoint difference \(\frac{13}{6}\); Primitive \(- \frac{13 x^{2}}{12} + \frac{26 x}{3}\), endpoint difference \(\frac{13}{3}\).
Therefore the area is \(\frac{13}{2}\) square units.

[1 mark] In increasing x order, the vertices are (1,2), (2,-2), (4,3). The left edge, right edge and long edge have equations y=\(6 - 4 x\), y=\(\frac{5 x}{2} - 7\), y=\(\frac{x}{3} + \frac{5}{3}\), respectively.
[1 mark] Use upper minus lower strip widths: \(\frac{13 x}{3} - \frac{13}{3}\) for x∈\(\left[1, 2\right]\); \(\frac{26}{3} - \frac{13 x}{6}\) for x∈\(\left[2, 4\right]\).
[1 mark] Hence A=\(\int_{1}^{2}(\frac{13 x}{3} - \frac{13}{3})\,dx\) + \(\int_{2}^{4}(\frac{26}{3} - \frac{13 x}{6})\,dx\).
[1 mark] Primitive \(\frac{13 x^{2}}{6} - \frac{13 x}{3}\), endpoint difference \(\frac{13}{6}\); Primitive \(- \frac{13 x^{2}}{12} + \frac{26 x}{3}\), endpoint difference \(\frac{13}{3}\).
[1 mark] Therefore the area is \(\frac{13}{2}\) square units.

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