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प्रश्न
Evaluate: `int_-2^1 sqrt(5 - 4x - x^2)dx`
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उत्तर
Let I = `int_-2^1 sqrt(5 - 4x - x^2)dx`
= `int_-2^1 sqrt(-(x^2 + 4x - 5))dx`
= `int_2^1 sqrt(-(x^2 + 4x + 2^2 - 2^2 - 5))dx`
= `int_-2^1 sqrt(-{(x + 2)^2 - 9})dx`
= `int_-2^1 sqrt(3^2 - (x + 2)^2)dx`
= `[(x + 2)/2 sqrt(3^2 - (x + 2)^2) + 3^2/2 sin^-1 ((x + 2)/3)]_-2^1`
= `0 + 9/2 . π/2 - (0 + 0)`
= `(9π)/4`
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