Class 12 Applications Of Integrals · Very Long Answer5 Marksarea_by_integration

Class 12 Maths · Applications Of Integrals · Very Long Answer #45

Sketch and find the area of {(x,y):0≤y≤x²+3,0≤y≤2x+3,0<x<3}.

✅ Step-by-Step Solution

Final Answer: \(\frac{50}{3}\) square units

The upper boundary is min(x²+3,2x+3). The two curves intersect at x=0,2; use the parabola for 0<x<2 and the line for 2<x<3. Omitting vertical endpoints does not change area.
Use upper minus lower strip widths: \(x^{2} + 3\) for x∈\(\left[0, 2\right]\); \(2 x + 3\) for x∈\(\left[2, 3\right]\).
Hence A=\(\int_{0}^{2}(x^{2} + 3)\,dx\) + \(\int_{2}^{3}(2 x + 3)\,dx\).
Primitive \(\frac{x^{3}}{3} + 3 x\), endpoint difference \(\frac{26}{3}\); Primitive \(x^{2} + 3 x\), endpoint difference \(8\).
Therefore the area is \(\frac{50}{3}\) square units.

[1 mark] The upper boundary is min(x²+3,2x+3). The two curves intersect at x=0,2; use the parabola for 0<x<2 and the line for 2<x<3. Omitting vertical endpoints does not change area.
[1 mark] Use upper minus lower strip widths: \(x^{2} + 3\) for x∈\(\left[0, 2\right]\); \(2 x + 3\) for x∈\(\left[2, 3\right]\).
[1 mark] Hence A=\(\int_{0}^{2}(x^{2} + 3)\,dx\) + \(\int_{2}^{3}(2 x + 3)\,dx\).
[1 mark] Primitive \(\frac{x^{3}}{3} + 3 x\), endpoint difference \(\frac{26}{3}\); Primitive \(x^{2} + 3 x\), endpoint difference \(8\).
[1 mark] Therefore the area is \(\frac{50}{3}\) square units.

Practise interactive questions, flashcards, visual mind maps, and timed board tests for Applications Of Integrals.