Class 12 Applications Of Integrals · Very Long Answer5 Marksarea_by_integration

Class 12 Maths · Applications Of Integrals · Very Long Answer #19

Sketch and find by integration the area of {(x,y):0≤y≤x²,0≤y≤x,0≤x≤2}.

✅ Step-by-Step Solution

Final Answer: \(\frac{11}{6}\) square units

The upper boundary is min(x²,x). Since x²=x at x=0,1, use the parabola on [0,1] and the line on [1,2].
Use upper minus lower strip widths: \(x^{2}\) for x∈\(\left[0, 1\right]\); \(x\) for x∈\(\left[1, 2\right]\).
Hence A=\(\int_{0}^{1}(x^{2})\,dx\) + \(\int_{1}^{2}(x)\,dx\).
Primitive \(\frac{x^{3}}{3}\), endpoint difference \(\frac{1}{3}\); Primitive \(\frac{x^{2}}{2}\), endpoint difference \(\frac{3}{2}\).
Therefore the area is \(\frac{11}{6}\) square units.

[1 mark] The upper boundary is min(x²,x). Since x²=x at x=0,1, use the parabola on [0,1] and the line on [1,2].
[1 mark] Use upper minus lower strip widths: \(x^{2}\) for x∈\(\left[0, 1\right]\); \(x\) for x∈\(\left[1, 2\right]\).
[1 mark] Hence A=\(\int_{0}^{1}(x^{2})\,dx\) + \(\int_{1}^{2}(x)\,dx\).
[1 mark] Primitive \(\frac{x^{3}}{3}\), endpoint difference \(\frac{1}{3}\); Primitive \(\frac{x^{2}}{2}\), endpoint difference \(\frac{3}{2}\).
[1 mark] Therefore the area is \(\frac{11}{6}\) square units.

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