Class 12 Applications Of Integrals · Very Long Answer5 Marksarea_by_integration

Class 12 Maths · Applications Of Integrals · Very Long Answer #28

Find the area enclosed between the parabola 4y=3x² and the line 3x−2y+12=0.

✅ Step-by-Step Solution

Final Answer: \(27\) square units

Intersections solve 3x²/4=3x/2+6, giving x=−2,4 and points (−2,3),(4,12). The line lies above the parabola between them.
Use upper minus lower strip widths: \(- \frac{3 x^{2}}{4} + \frac{3 x}{2} + 6\) for x∈\(\left[-2, 4\right]\).
Hence A=\(\int_{-2}^{4}(- \frac{3 x^{2}}{4} + \frac{3 x}{2} + 6)\,dx\).
Primitive \(- \frac{x^{3}}{4} + \frac{3 x^{2}}{4} + 6 x\), endpoint difference \(27\).
Therefore the area is \(27\) square units.

[1 mark] Intersections solve 3x²/4=3x/2+6, giving x=−2,4 and points (−2,3),(4,12). The line lies above the parabola between them.
[1 mark] Use upper minus lower strip widths: \(- \frac{3 x^{2}}{4} + \frac{3 x}{2} + 6\) for x∈\(\left[-2, 4\right]\).
[1 mark] Hence A=\(\int_{-2}^{4}(- \frac{3 x^{2}}{4} + \frac{3 x}{2} + 6)\,dx\).
[1 mark] Primitive \(- \frac{x^{3}}{4} + \frac{3 x^{2}}{4} + 6 x\), endpoint difference \(27\).
[1 mark] Therefore the area is \(27\) square units.

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