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प्रश्न
Evaluate the definite integral:
`int_(-1)^1 (x + 1)dx`
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उत्तर
`int_-1^1 (x + 1) dx = [x^2/2 + x]_-1^1`
`= 1/2 [(1)^2 - (-1)^2] + [1 - (-1)]`
`= 1/2 (1 - 1) + (1 + 1)`
`= 1/2 (0) + 2`
= 2
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