Class 12 Applications Of Integrals · Very Long Answer5 Marksarea_by_integration

Class 12 Maths · Applications Of Integrals · Very Long Answer #51

Using integration, find the area of the triangle with vertices (2,3), (3,5), (4,4).

✅ Step-by-Step Solution

Final Answer: \(\frac{3}{2}\) square units

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

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

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