Class 12 Applications Of Integrals · Short Answer2 Marksarea_by_integration
Class 12 Maths · Applications Of Integrals · Short Answer #11
For a>0, find the area under y=√(a²−x²), above the x-axis, between x=0 and x=a.
✅ Step-by-Step Solution
Final Answer: \(\frac{\pi a^{2}}{4}\) square units
- This is the first-quadrant quarter disk of radius a. A vertical strip has height √(a²−x²). The strip widths are \(\sqrt{a^{2} - x^{2}}\) for x∈\(\left[0, a\right]\).
- Primitive \(\frac{a^{2} \operatorname{asin}{\left(\frac{x}{a} \right)}}{2} + \frac{x \sqrt{a^{2} - x^{2}}}{2}\), endpoint difference \(\frac{\pi a^{2}}{4}\).
- Therefore the area is \(\frac{\pi a^{2}}{4}\) square units.
📋 CBSE Stepwise Marking Scheme
- [0.5 marks] This is the first-quadrant quarter disk of radius a. A vertical strip has height √(a²−x²). The strip widths are \(\sqrt{a^{2} - x^{2}}\) for x∈\(\left[0, a\right]\).
- [1 mark] Primitive \(\frac{a^{2} \operatorname{asin}{\left(\frac{x}{a} \right)}}{2} + \frac{x \sqrt{a^{2} - x^{2}}}{2}\), endpoint difference \(\frac{\pi a^{2}}{4}\).
- [0.5 marks] Therefore the area is \(\frac{\pi a^{2}}{4}\) square units.
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