Class 12 Applications Of Integrals · Very Long Answer5 Marksarea_by_integration

Class 12 Maths · Applications Of Integrals · Very Long Answer #41

Find by integration the area enclosed between x²=4y and x=4y−2.

✅ Step-by-Step Solution

Final Answer: \(\frac{9}{8}\) square units

The line is y=(x+2)/4. Intersections solve x²=x+2, hence x=−1,2; the line lies above the parabola between these inputs.
Use upper minus lower strip widths: \(- \frac{x^{2}}{4} + \frac{x}{4} + \frac{1}{2}\) for x∈\(\left[-1, 2\right]\).
Hence A=\(\int_{-1}^{2}(- \frac{x^{2}}{4} + \frac{x}{4} + \frac{1}{2})\,dx\).
Primitive \(- \frac{x^{3}}{12} + \frac{x^{2}}{8} + \frac{x}{2}\), endpoint difference \(\frac{9}{8}\).
Therefore the area is \(\frac{9}{8}\) square units.

[1 mark] The line is y=(x+2)/4. Intersections solve x²=x+2, hence x=−1,2; the line lies above the parabola between these inputs.
[1 mark] Use upper minus lower strip widths: \(- \frac{x^{2}}{4} + \frac{x}{4} + \frac{1}{2}\) for x∈\(\left[-1, 2\right]\).
[1 mark] Hence A=\(\int_{-1}^{2}(- \frac{x^{2}}{4} + \frac{x}{4} + \frac{1}{2})\,dx\).
[1 mark] Primitive \(- \frac{x^{3}}{12} + \frac{x^{2}}{8} + \frac{x}{2}\), endpoint difference \(\frac{9}{8}\).
[1 mark] Therefore the area is \(\frac{9}{8}\) square units.

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