Class 12 Applications Of Integrals · Very Long Answer5 Marksarea_by_integration

Class 12 Maths · Applications Of Integrals · Very Long Answer #17

Using integration, find the first-quadrant area bounded by y=√3x, y=√(4−x²) and the y-axis.

✅ Step-by-Step Solution

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

The first-quadrant intersection solves 3x²=4−x², giving (1,√3). For 0≤x≤1 the circular arc lies above the line.
Use upper minus lower strip widths: \(- \sqrt{3} x + \sqrt{4 - x^{2}}\) for x∈\(\left[0, 1\right]\).
Hence A=\(\int_{0}^{1}(- \sqrt{3} x + \sqrt{4 - x^{2}})\,dx\).
Primitive \(- \frac{\sqrt{3} x^{2}}{2} + \frac{x \sqrt{4 - x^{2}}}{2} + 2 \operatorname{asin}{\left(\frac{x}{2} \right)}\), endpoint difference \(\frac{\pi}{3}\).
Therefore the area is \(\frac{\pi}{3}\) square units.

[1 mark] The first-quadrant intersection solves 3x²=4−x², giving (1,√3). For 0≤x≤1 the circular arc lies above the line.
[1 mark] Use upper minus lower strip widths: \(- \sqrt{3} x + \sqrt{4 - x^{2}}\) for x∈\(\left[0, 1\right]\).
[1 mark] Hence A=\(\int_{0}^{1}(- \sqrt{3} x + \sqrt{4 - x^{2}})\,dx\).
[1 mark] Primitive \(- \frac{\sqrt{3} x^{2}}{2} + \frac{x \sqrt{4 - x^{2}}}{2} + 2 \operatorname{asin}{\left(\frac{x}{2} \right)}\), endpoint difference \(\frac{\pi}{3}\).
[1 mark] Therefore the area is \(\frac{\pi}{3}\) square units.

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