Get the solution to "Evaluate the following integrals. ∫−2−1−4x−x2dx"

York

Answered question

2021-02-12

Evaluate the following integrals.

\int_{- 2}^{- 1} \sqrt{- 4 x - x^{2}} d x

yagombyeR

Skilled2021-02-13Added 92 answers

Step 1: Given that

\int_{- 2}^{- 1} \sqrt{- 4 x - x^{2}} d x

Step 2: Formula Used

\int \sqrt{a^{2} - x^{2}} = \frac{x}{2} \sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2} \sin^{- 1} (\frac{x}{a}) + C

Step 3: Solve
We have,

\int_{- 2}^{- 1} \sqrt{- 4 x - x^{2}} d x = \int_{- 2}^{- 1} \sqrt{- (4 x + x^{2})} d x

= \int_{- 2}^{- 1} \sqrt{- (4 x + x^{2} + 4 - 4)} d x

= \int_{- 2}^{- 1} \sqrt{- ({(x + 2)}^{2} - 4)} d x

= \int_{- 2}^{- 1} \sqrt{4 - {(x + 2)}^{2}} d x

= \int_{- 2}^{- 1} \sqrt{{(2)}^{2} - {(x + 2)}^{2}} d x

= {[\frac{x + 2}{2} \sqrt{4 - {(x + 2)}^{2}} + \frac{4}{2} \sin^{- 1} (\frac{x + 2}{2})]}_{- 2}^{- 1}

= [\frac{- 1 + 2}{2} \sqrt{4 - {(- 1 + 2)}^{2}} + 2 \sin^{- 1} (\frac{- 1 + 2}{2}) - \frac{- 2 + 2}{2} \sqrt{4 - {(- 2 + 2)}^{2}} = 2 \sin^{- 1} (\frac{- 2 + 2}{2})]

= [\frac{1}{2} \sqrt{4 - 1} + 2 \sin^{- 1} (\frac{1}{2}) - 0 - 0]

= \frac{\sqrt{3}}{2} + 2 \sin^{- 1} (\sin \frac{π}{6})

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