Class 12 Applications Of Integrals · Very Long Answer5 Marksarea_by_integration

Class 12 Maths · Applications Of Integrals · Very Long Answer #22

Using integration, find the area of the triangle with vertices (2,-2), (4,5), (6,2).

✅ Step-by-Step Solution

Final Answer: \(10\) square units

In increasing x order, the vertices are (2,-2), (4,5), (6,2). The left edge, right edge and long edge have equations y=\(\frac{7 x}{2} - 9\), y=\(11 - \frac{3 x}{2}\), y=\(x - 4\), respectively.
Use upper minus lower strip widths: \(\frac{5 x}{2} - 5\) for x∈\(\left[2, 4\right]\); \(15 - \frac{5 x}{2}\) for x∈\(\left[4, 6\right]\).
Hence A=\(\int_{2}^{4}(\frac{5 x}{2} - 5)\,dx\) + \(\int_{4}^{6}(15 - \frac{5 x}{2})\,dx\).
Primitive \(\frac{5 x^{2}}{4} - 5 x\), endpoint difference \(5\); Primitive \(- \frac{5 x^{2}}{4} + 15 x\), endpoint difference \(5\).
Therefore the area is \(10\) square units.

[1 mark] In increasing x order, the vertices are (2,-2), (4,5), (6,2). The left edge, right edge and long edge have equations y=\(\frac{7 x}{2} - 9\), y=\(11 - \frac{3 x}{2}\), y=\(x - 4\), respectively.
[1 mark] Use upper minus lower strip widths: \(\frac{5 x}{2} - 5\) for x∈\(\left[2, 4\right]\); \(15 - \frac{5 x}{2}\) for x∈\(\left[4, 6\right]\).
[1 mark] Hence A=\(\int_{2}^{4}(\frac{5 x}{2} - 5)\,dx\) + \(\int_{4}^{6}(15 - \frac{5 x}{2})\,dx\).
[1 mark] Primitive \(\frac{5 x^{2}}{4} - 5 x\), endpoint difference \(5\); Primitive \(- \frac{5 x^{2}}{4} + 15 x\), endpoint difference \(5\).
[1 mark] Therefore the area is \(10\) square units.

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