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Evaluate: ∫-215-4x-x2dx

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प्रश्न

Evaluate: `int_-2^1 sqrt(5 - 4x - x^2)dx`

योग
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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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2021-2022 (March) Term 2 - Outside Delhi Set 1

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