Class 12 Applications Of Integrals · Very Long Answer5 Marksarea_by_integration

Class 12 Maths · Applications Of Integrals · Very Long Answer #24

Find by integration the area of {(x,y):x²+y²≤4,x+y≥2}.

✅ Step-by-Step Solution

Final Answer: \(-2 + \pi\) square units

The circle–line intersections are (0,2),(2,0). Between them the circular arc is above y=2−x.
Use upper minus lower strip widths: \(x + \sqrt{4 - x^{2}} - 2\) for x∈\(\left[0, 2\right]\).
Hence A=\(\int_{0}^{2}(x + \sqrt{4 - x^{2}} - 2)\,dx\).
Primitive \(\frac{x^{2}}{2} + \frac{x \sqrt{4 - x^{2}}}{2} - 2 x + 2 \operatorname{asin}{\left(\frac{x}{2} \right)}\), endpoint difference \(-2 + \pi\).
Therefore the area is \(-2 + \pi\) square units.

[1 mark] The circle–line intersections are (0,2),(2,0). Between them the circular arc is above y=2−x.
[1 mark] Use upper minus lower strip widths: \(x + \sqrt{4 - x^{2}} - 2\) for x∈\(\left[0, 2\right]\).
[1 mark] Hence A=\(\int_{0}^{2}(x + \sqrt{4 - x^{2}} - 2)\,dx\).
[1 mark] Primitive \(\frac{x^{2}}{2} + \frac{x \sqrt{4 - x^{2}}}{2} - 2 x + 2 \operatorname{asin}{\left(\frac{x}{2} \right)}\), endpoint difference \(-2 + \pi\).
[1 mark] Therefore the area is \(-2 + \pi\) square units.

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