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If ∫ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ 2 0 2 x – x 2 – x 2 dx = ∫ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ 1 0 2 2 2 y – y – 1 – 1 dy + ∫ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ 2 1 2 2 y – 2 dy + I the I equal is

1. ∫ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + 1 0 2 dy y – 1 1
2. ∫ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + 1 0 2 2 dy 1 y – 1 – 2 y
3. ∫ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ 1 0 2 dy y – 1 – 1
4. ∫ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + + 1 0 2 2 dy 1 y – 1 2 y

Correct answer: ∫ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ 1 0 2 dy y – 1 – 1

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