Class 12 Applications Of Integrals · Very Long Answer5 Marksarea_by_integration
Class 12 Maths · Applications Of Integrals · Very Long Answer #18
Find using integration the area of {(x,y): y²≤2x and y≥x−4}.
✅ Step-by-Step Solution
Final Answer: \(18\) square units
- The inequalities give y²/2≤x≤y+4. Intersections satisfy y²/2=y+4, hence y=−2,4, at (2,−2),(8,4).
- Use right minus left strip widths: \(- \frac{y^{2}}{2} + y + 4\) for y∈\(\left[-2, 4\right]\).
- Hence A=\(\int_{-2}^{4}(- \frac{y^{2}}{2} + y + 4)\,dy\).
- Primitive \(- \frac{y^{3}}{6} + \frac{y^{2}}{2} + 4 y\), endpoint difference \(18\).
- Therefore the area is \(18\) square units.
📋 CBSE Stepwise Marking Scheme
- [1 mark] The inequalities give y²/2≤x≤y+4. Intersections satisfy y²/2=y+4, hence y=−2,4, at (2,−2),(8,4).
- [1 mark] Use right minus left strip widths: \(- \frac{y^{2}}{2} + y + 4\) for y∈\(\left[-2, 4\right]\).
- [1 mark] Hence A=\(\int_{-2}^{4}(- \frac{y^{2}}{2} + y + 4)\,dy\).
- [1 mark] Primitive \(- \frac{y^{3}}{6} + \frac{y^{2}}{2} + 4 y\), endpoint difference \(18\).
- [1 mark] Therefore the area is \(18\) square units.
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