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Class 12 Applications Of Integrals · Very Long Answer5 Marksarea_by_integration

Class 12 Maths · Applications Of Integrals · Very Long Answer #16

Using integration, find the area of the small region bounded by y=x², y=x+2 and the x-axis with −2≤x≤0. The line segment runs from (−2,0) to (−1,1), and the parabola segment from (−1,1) to (0,0).
Computed region for la 2
Computed region for la 2

✅ Step-by-Step Solution

Final Answer: \(\frac{5}{6}\) square units
  1. The line crosses the x-axis at (−2,0); the parabola touches it at (0,0). The relevant line–parabola intersection is (−1,1), from x²=x+2.
  2. Use upper minus lower strip widths: \(x + 2\) for x∈\(\left[-2, -1\right]\); \(x^{2}\) for x∈\(\left[-1, 0\right]\).
  3. Hence A=\(\int_{-2}^{-1}(x + 2)\,dx\) + \(\int_{-1}^{0}(x^{2})\,dx\).
  4. Primitive \(\frac{x^{2}}{2} + 2 x\), endpoint difference \(\frac{1}{2}\); Primitive \(\frac{x^{3}}{3}\), endpoint difference \(\frac{1}{3}\).
  5. Therefore the area is \(\frac{5}{6}\) square units.

📋 CBSE Stepwise Marking Scheme

  • [1 mark] The line crosses the x-axis at (−2,0); the parabola touches it at (0,0). The relevant line–parabola intersection is (−1,1), from x²=x+2.
  • [1 mark] Use upper minus lower strip widths: \(x + 2\) for x∈\(\left[-2, -1\right]\); \(x^{2}\) for x∈\(\left[-1, 0\right]\).
  • [1 mark] Hence A=\(\int_{-2}^{-1}(x + 2)\,dx\) + \(\int_{-1}^{0}(x^{2})\,dx\).
  • [1 mark] Primitive \(\frac{x^{2}}{2} + 2 x\), endpoint difference \(\frac{1}{2}\); Primitive \(\frac{x^{3}}{3}\), endpoint difference \(\frac{1}{3}\).
  • [1 mark] Therefore the area is \(\frac{5}{6}\) square units.

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