Class 12 Applications Of Integrals · Very Long Answer5 Marksarea_by_integration

Class 12 Maths · Applications Of Integrals · Very Long Answer #52

Using integration, find the area of the triangle with vertices (-1,0), (1,3), (3,2).

✅ Step-by-Step Solution

Final Answer: \(4\) square units

In increasing x order, the vertices are (-1,0), (1,3), (3,2). The left edge, right edge and long edge have equations y=\(\frac{3 x}{2} + \frac{3}{2}\), y=\(\frac{7}{2} - \frac{x}{2}\), y=\(\frac{x}{2} + \frac{1}{2}\), respectively.
Use upper minus lower strip widths: \(x + 1\) for x∈\(\left[-1, 1\right]\); \(3 - x\) for x∈\(\left[1, 3\right]\).
Hence A=\(\int_{-1}^{1}(x + 1)\,dx\) + \(\int_{1}^{3}(3 - x)\,dx\).
Primitive \(\frac{x^{2}}{2} + x\), endpoint difference \(2\); Primitive \(- \frac{x^{2}}{2} + 3 x\), endpoint difference \(2\).
Therefore the area is \(4\) square units.

[1 mark] In increasing x order, the vertices are (-1,0), (1,3), (3,2). The left edge, right edge and long edge have equations y=\(\frac{3 x}{2} + \frac{3}{2}\), y=\(\frac{7}{2} - \frac{x}{2}\), y=\(\frac{x}{2} + \frac{1}{2}\), respectively.
[1 mark] Use upper minus lower strip widths: \(x + 1\) for x∈\(\left[-1, 1\right]\); \(3 - x\) for x∈\(\left[1, 3\right]\).
[1 mark] Hence A=\(\int_{-1}^{1}(x + 1)\,dx\) + \(\int_{1}^{3}(3 - x)\,dx\).
[1 mark] Primitive \(\frac{x^{2}}{2} + x\), endpoint difference \(2\); Primitive \(- \frac{x^{2}}{2} + 3 x\), endpoint difference \(2\).
[1 mark] Therefore the area is \(4\) square units.

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