Class 12 Applications Of Integrals · Very Long Answer5 Marksarea_by_integration

Class 12 Maths · Applications Of Integrals · Very Long Answer #40

Using integration, find the area enclosed by y=4x+5,y=5−x and 4y=x+5.

✅ Step-by-Step Solution

Final Answer: \(\frac{15}{2}\) square units

In increasing x order, the vertices are (-1,1), (0,5), (3,2). The left edge, right edge and long edge have equations y=\(4 x + 5\), y=\(5 - x\), y=\(\frac{x}{4} + \frac{5}{4}\), respectively.
Use upper minus lower strip widths: \(\frac{15 x}{4} + \frac{15}{4}\) for x∈\(\left[-1, 0\right]\); \(\frac{15}{4} - \frac{5 x}{4}\) for x∈\(\left[0, 3\right]\).
Hence A=\(\int_{-1}^{0}(\frac{15 x}{4} + \frac{15}{4})\,dx\) + \(\int_{0}^{3}(\frac{15}{4} - \frac{5 x}{4})\,dx\).
Primitive \(\frac{15 x^{2}}{8} + \frac{15 x}{4}\), endpoint difference \(\frac{15}{8}\); Primitive \(- \frac{5 x^{2}}{8} + \frac{15 x}{4}\), endpoint difference \(\frac{45}{8}\).
Therefore the area is \(\frac{15}{2}\) square units.

[1 mark] In increasing x order, the vertices are (-1,1), (0,5), (3,2). The left edge, right edge and long edge have equations y=\(4 x + 5\), y=\(5 - x\), y=\(\frac{x}{4} + \frac{5}{4}\), respectively.
[1 mark] Use upper minus lower strip widths: \(\frac{15 x}{4} + \frac{15}{4}\) for x∈\(\left[-1, 0\right]\); \(\frac{15}{4} - \frac{5 x}{4}\) for x∈\(\left[0, 3\right]\).
[1 mark] Hence A=\(\int_{-1}^{0}(\frac{15 x}{4} + \frac{15}{4})\,dx\) + \(\int_{0}^{3}(\frac{15}{4} - \frac{5 x}{4})\,dx\).
[1 mark] Primitive \(\frac{15 x^{2}}{8} + \frac{15 x}{4}\), endpoint difference \(\frac{15}{8}\); Primitive \(- \frac{5 x^{2}}{8} + \frac{15 x}{4}\), endpoint difference \(\frac{45}{8}\).
[1 mark] Therefore the area is \(\frac{15}{2}\) square units.

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