Class 12 Applications Of Integrals · Very Long Answer5 Marksarea_by_integration

Class 12 Maths · Applications Of Integrals · Very Long Answer #30

Sketch y=|x+1|, evaluate ∫₋₃¹|x+1|dx, and state the geometric meaning of the result.

✅ Step-by-Step Solution

Final Answer: \(4\) square units

The V-graph has vertex (−1,0), with y=−x−1 for x<−1 and y=x+1 for x≥−1. The integral is the nonnegative area above the x-axis between x=−3 and x=1.
Use upper minus lower strip widths: \(- x - 1\) for x∈\(\left[-3, -1\right]\); \(x + 1\) for x∈\(\left[-1, 1\right]\).
Hence A=\(\int_{-3}^{-1}(- x - 1)\,dx\) + \(\int_{-1}^{1}(x + 1)\,dx\).
Primitive \(- \frac{x^{2}}{2} - x\), endpoint difference \(2\); Primitive \(\frac{x^{2}}{2} + x\), endpoint difference \(2\).
Therefore the area is \(4\) square units.

[1 mark] The V-graph has vertex (−1,0), with y=−x−1 for x<−1 and y=x+1 for x≥−1. The integral is the nonnegative area above the x-axis between x=−3 and x=1.
[1 mark] Use upper minus lower strip widths: \(- x - 1\) for x∈\(\left[-3, -1\right]\); \(x + 1\) for x∈\(\left[-1, 1\right]\).
[1 mark] Hence A=\(\int_{-3}^{-1}(- x - 1)\,dx\) + \(\int_{-1}^{1}(x + 1)\,dx\).
[1 mark] Primitive \(- \frac{x^{2}}{2} - x\), endpoint difference \(2\); Primitive \(\frac{x^{2}}{2} + x\), endpoint difference \(2\).
[1 mark] Therefore the area is \(4\) square units.

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