Class 12 Applications Of Integrals · Short Answer2 Marksarea_by_integration
Class 12 Maths · Applications Of Integrals · Short Answer #39
Find the area under y=√(x−1) above the x-axis between x=1 and x=5.
✅ Step-by-Step Solution
Final Answer: \(\frac{16}{3}\) square units
- The curve is nonnegative on [1,5], beginning at (1,0). Use its height √(x−1). The strip widths are \(\sqrt{x - 1}\) for x∈\(\left[1, 5\right]\).
- Primitive \(\frac{2 \left(x - 1\right)^{\frac{3}{2}}}{3}\), endpoint difference \(\frac{16}{3}\).
- Therefore the area is \(\frac{16}{3}\) square units.
📋 CBSE Stepwise Marking Scheme
- [0.5 marks] The curve is nonnegative on [1,5], beginning at (1,0). Use its height √(x−1). The strip widths are \(\sqrt{x - 1}\) for x∈\(\left[1, 5\right]\).
- [1 mark] Primitive \(\frac{2 \left(x - 1\right)^{\frac{3}{2}}}{3}\), endpoint difference \(\frac{16}{3}\).
- [0.5 marks] Therefore the area is \(\frac{16}{3}\) square units.
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