York

2021-02-12

Evaluate the following integrals.
${\int }_{-2}^{-1}\sqrt{-4x-{x}^{2}}dx$

yagombyeR

Step 1: Given that
${\int }_{-2}^{-1}\sqrt{-4x-{x}^{2}}dx$
Step 2: Formula Used
$\int \sqrt{{a}^{2}-{x}^{2}}=\frac{x}{2}\sqrt{{a}^{2}-{x}^{2}}+\frac{{a}^{2}}{2}{\mathrm{sin}}^{-1}\left(\frac{x}{a}\right)+C$
Step 3: Solve
We have,
${\int }_{-2}^{-1}\sqrt{-4x-{x}^{2}}dx={\int }_{-2}^{-1}\sqrt{-\left(4x+{x}^{2}\right)}dx$
$={\int }_{-2}^{-1}\sqrt{-\left(4x+{x}^{2}+4-4\right)}dx$
$={\int }_{-2}^{-1}\sqrt{-\left({\left(x+2\right)}^{2}-4\right)}dx$
$={\int }_{-2}^{-1}\sqrt{4-{\left(x+2\right)}^{2}}dx$
$={\int }_{-2}^{-1}\sqrt{{\left(2\right)}^{2}-{\left(x+2\right)}^{2}}dx$
$={\left[\frac{x+2}{2}\sqrt{4-{\left(x+2\right)}^{2}}+\frac{4}{2}{\mathrm{sin}}^{-1}\left(\frac{x+2}{2}\right)\right]}_{-2}^{-1}$
$=\left[\frac{-1+2}{2}\sqrt{4-{\left(-1+2\right)}^{2}}+2{\mathrm{sin}}^{-1}\left(\frac{-1+2}{2}\right)-\frac{-2+2}{2}\sqrt{4-{\left(-2+2\right)}^{2}}=2{\mathrm{sin}}^{-1}\left(\frac{-2+2}{2}\right)\right]$
$=\left[\frac{1}{2}\sqrt{4-1}+2{\mathrm{sin}}^{-1}\left(\frac{1}{2}\right)-0-0\right]$
$=\frac{\sqrt{3}}{2}+2{\mathrm{sin}}^{-1}\left(\mathrm{sin}\frac{\pi }{6}\right)$

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