Answer
8
Work Step by Step
\[\begin{gathered}
\int_1^9 {\frac{2}{{\sqrt x }}} dx \hfill \\
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{\text{Use }}\sqrt x = {x^{1/2}} \hfill \\
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\int_1^9 {\frac{2}{{{x^{1/2}}}}} dx \hfill \\
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{\text{Use }}\frac{1}{{{x^{1/2}}}} = {x^{ - 1/2}} \hfill \\
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\int_1^9 {\frac{2}{{{x^{1/2}}}}} dx = \int_1^9 {2{x^{ - 1/2}}} dx \hfill \\
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{\text{Use }}\int {{x^n}dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C{\text{ }}} \hfill \\
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\int_1^9 {2{x^{ - 1/2}}} dx = 2\left[ {\frac{{{x^{ - 1/2 + 1}}}}{{ - 1/2 + 1}}} \right]_1^9 \hfill \\
\hfill \\
= 2\left( {\frac{{{x^{1/2}}}}{{1/2}}} \right)_1^9 = 4\left( {{x^{1/2}}} \right)_1^9 \hfill \\
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{\text{Fundamental Theorem of calculus}} \hfill \\
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4\left( {{{\left( 9 \right)}^{1/2}} - {{\left( 1 \right)}^{1/2}}} \right) \hfill \\
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{\text{Simplify}} \hfill \\
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4\left( 2 \right) \hfill \\
\hfill \\
8 \hfill \\
\end{gathered} \]
