Class 12 Applications Of Integrals · Short Answer2 Marksarea_by_integration
Class 12 Maths · Applications Of Integrals · Short Answer #38
Find the area under y=2√x above the x-axis from x=0 to x=1.
✅ Step-by-Step Solution
Final Answer: \(\frac{4}{3}\) square units
- The real curve starts at (0,0) and is nonnegative; the given x-limits are 0 and 1. The strip widths are \(2 \sqrt{x}\) for x∈\(\left[0, 1\right]\).
- Primitive \(\frac{4 x^{\frac{3}{2}}}{3}\), endpoint difference \(\frac{4}{3}\).
- Therefore the area is \(\frac{4}{3}\) square units.
📋 CBSE Stepwise Marking Scheme
- [0.5 marks] The real curve starts at (0,0) and is nonnegative; the given x-limits are 0 and 1. The strip widths are \(2 \sqrt{x}\) for x∈\(\left[0, 1\right]\).
- [1 mark] Primitive \(\frac{4 x^{\frac{3}{2}}}{3}\), endpoint difference \(\frac{4}{3}\).
- [0.5 marks] Therefore the area is \(\frac{4}{3}\) square units.
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