Class 12 Applications Of Integrals · Very Long Answer5 Marksarea_by_integration

Class 12 Maths · Applications Of Integrals · Very Long Answer #29

Using integration, find the area bounded by the tangent to 4y=x² at (2,1), and the lines x=2y and x=3y−3.

✅ Step-by-Step Solution

Final Answer: \(1\) square units

Differentiating 4y=x² gives y′=x/2, hence tangent y−1=x−2 or y=x−1. Its intersections with y=x/2 and y=(x+3)/3 are (2,1),(3,2); the latter two lines meet at (6,3). The three boundary lines are y=x−1,y=(x+3)/3,y=x/2.
Use upper minus lower strip widths: \(\frac{x}{2} - 1\) for x∈\(\left[2, 3\right]\); \(1 - \frac{x}{6}\) for x∈\(\left[3, 6\right]\).
Hence A=\(\int_{2}^{3}(\frac{x}{2} - 1)\,dx\) + \(\int_{3}^{6}(1 - \frac{x}{6})\,dx\).
Primitive \(\frac{x^{2}}{4} - x\), endpoint difference \(\frac{1}{4}\); Primitive \(- \frac{x^{2}}{12} + x\), endpoint difference \(\frac{3}{4}\).
Therefore the area is \(1\) square units.

[1 mark] Differentiating 4y=x² gives y′=x/2, hence tangent y−1=x−2 or y=x−1. Its intersections with y=x/2 and y=(x+3)/3 are (2,1),(3,2); the latter two lines meet at (6,3). The three boundary lines are y=x−1,y=(x+3)/3,y=x/2.
[1 mark] Use upper minus lower strip widths: \(\frac{x}{2} - 1\) for x∈\(\left[2, 3\right]\); \(1 - \frac{x}{6}\) for x∈\(\left[3, 6\right]\).
[1 mark] Hence A=\(\int_{2}^{3}(\frac{x}{2} - 1)\,dx\) + \(\int_{3}^{6}(1 - \frac{x}{6})\,dx\).
[1 mark] Primitive \(\frac{x^{2}}{4} - x\), endpoint difference \(\frac{1}{4}\); Primitive \(- \frac{x^{2}}{12} + x\), endpoint difference \(\frac{3}{4}\).
[1 mark] Therefore the area is \(1\) square units.

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