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

Question
Applications of integrals
asked 2021-02-12
Evaluate the following integrals.
\(\displaystyle{\int_{{-{2}}}^{{-{1}}}}\sqrt{{-{4}{x}-{x}^{{{2}}}}}{\left.{d}{x}\right.}\)

Answers (1)

2021-02-13
Step 1: Given that
\(\displaystyle{\int_{{-{2}}}^{{-{1}}}}\sqrt{{-{4}{x}-{x}^{{{2}}}}}{\left.{d}{x}\right.}\)
Step 2: Formula Used
\(\displaystyle\int\sqrt{{{a}^{{{2}}}-{x}^{{{2}}}}}={\frac{{{x}}}{{{2}}}}\sqrt{{{a}^{{{2}}}-{x}^{{{2}}}}}+{\frac{{{a}^{{{2}}}}}{{{2}}}}{{\sin}^{{-{1}}}{\left({\frac{{{x}}}{{{a}}}}\right)}}+{C}\)
Step 3: Solve
We have,
\(\displaystyle{\int_{{-{2}}}^{{-{1}}}}\sqrt{{-{4}{x}-{x}^{{{2}}}}}{\left.{d}{x}\right.}={\int_{{-{2}}}^{{-{1}}}}\sqrt{{-{\left({4}{x}+{x}^{{{2}}}\right)}}}{\left.{d}{x}\right.}\)
\(\displaystyle={\int_{{-{2}}}^{{-{1}}}}\sqrt{{-{\left({4}{x}+{x}^{{{2}}}+{4}-{4}\right)}}}{\left.{d}{x}\right.}\)
\(\displaystyle={\int_{{-{2}}}^{{-{1}}}}\sqrt{{-{\left({\left({x}+{2}\right)}^{{{2}}}-{4}\right)}}}{\left.{d}{x}\right.}\)
\(\displaystyle={\int_{{-{2}}}^{{-{1}}}}\sqrt{{{4}-{\left({x}+{2}\right)}^{{{2}}}}}{\left.{d}{x}\right.}\)
\(\displaystyle={\int_{{-{2}}}^{{-{1}}}}\sqrt{{{\left({2}\right)}^{{{2}}}-{\left({x}+{2}\right)}^{{{2}}}}}{\left.{d}{x}\right.}\)
\(\displaystyle={{\left[{\frac{{{x}+{2}}}{{{2}}}}\sqrt{{{4}-{\left({x}+{2}\right)}^{{{2}}}}}+{\frac{{{4}}}{{{2}}}}{{\sin}^{{-{1}}}{\left({\frac{{{x}+{2}}}{{{2}}}}\right)}}\right]}_{{-{2}}}^{{-{1}}}}\)
\(\displaystyle={\left[{\frac{{-{1}+{2}}}{{{2}}}}\sqrt{{{4}-{\left(-{1}+{2}\right)}^{{{2}}}}}+{2}{{\sin}^{{-{1}}}{\left({\frac{{-{1}+{2}}}{{{2}}}}\right)}}-{\frac{{-{2}+{2}}}{{{2}}}}\sqrt{{{4}-{\left(-{2}+{2}\right)}^{{{2}}}}}={2}{{\sin}^{{-{1}}}{\left({\frac{{-{2}+{2}}}{{{2}}}}\right)}}\right]}\)
\(\displaystyle={\left[{\frac{{{1}}}{{{2}}}}\sqrt{{{4}-{1}}}+{2}{{\sin}^{{-{1}}}{\left({\frac{{{1}}}{{{2}}}}\right)}}-{0}-{0}\right]}\)
\(\displaystyle={\frac{{\sqrt{{{3}}}}}{{{2}}}}+{2}{{\sin}^{{-{1}}}{\left({\sin{{\frac{{\pi}}{{{6}}}}}}\right)}}\)
\(\displaystyle={\frac{{\sqrt{{{3}}}}}{{{2}}}}+{2}\times{\frac{{\pi}}{{{6}}}}\)
\(\displaystyle={\frac{{\sqrt{{{3}}}}}{{{2}}}}+{\frac{{\pi}}{{{3}}}}\)
Step 4: Result
\(\displaystyle{\int_{{-{2}}}^{{-{1}}}}\sqrt{{-{4}{x}-{x}^{{{2}}}}}{\left.{d}{x}\right.}={\frac{{\sqrt{{{3}}}}}{{{2}}}}+{\frac{{\pi}}{{{3}}}}\)
0

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