Class 12 Applications Of Integrals · Very Long Answer5 Marksarea_by_integration

Class 12 Maths · Applications Of Integrals · Very Long Answer #25

Using integration, find the area of the triangle with vertices (4,1), (6,6), (8,4).

✅ Step-by-Step Solution

Final Answer: \(7\) square units

In increasing x order, the vertices are (4,1), (6,6), (8,4). The left edge, right edge and long edge have equations y=\(\frac{5 x}{2} - 9\), y=\(12 - x\), y=\(\frac{3 x}{4} - 2\), respectively.
Use upper minus lower strip widths: \(\frac{7 x}{4} - 7\) for x∈\(\left[4, 6\right]\); \(14 - \frac{7 x}{4}\) for x∈\(\left[6, 8\right]\).
Hence A=\(\int_{4}^{6}(\frac{7 x}{4} - 7)\,dx\) + \(\int_{6}^{8}(14 - \frac{7 x}{4})\,dx\).
Primitive \(\frac{7 x^{2}}{8} - 7 x\), endpoint difference \(\frac{7}{2}\); Primitive \(- \frac{7 x^{2}}{8} + 14 x\), endpoint difference \(\frac{7}{2}\).
Therefore the area is \(7\) square units.

[1 mark] In increasing x order, the vertices are (4,1), (6,6), (8,4). The left edge, right edge and long edge have equations y=\(\frac{5 x}{2} - 9\), y=\(12 - x\), y=\(\frac{3 x}{4} - 2\), respectively.
[1 mark] Use upper minus lower strip widths: \(\frac{7 x}{4} - 7\) for x∈\(\left[4, 6\right]\); \(14 - \frac{7 x}{4}\) for x∈\(\left[6, 8\right]\).
[1 mark] Hence A=\(\int_{4}^{6}(\frac{7 x}{4} - 7)\,dx\) + \(\int_{6}^{8}(14 - \frac{7 x}{4})\,dx\).
[1 mark] Primitive \(\frac{7 x^{2}}{8} - 7 x\), endpoint difference \(\frac{7}{2}\); Primitive \(- \frac{7 x^{2}}{8} + 14 x\), endpoint difference \(\frac{7}{2}\).
[1 mark] Therefore the area is \(7\) square units.

Practise interactive questions, flashcards, visual mind maps, and timed board tests for Applications Of Integrals.