Find the following integral. \int \frac{3-x}{\sqrt{x}}dx

Find the following integral. \int \frac{3-x}{\sqrt{x}}dx

Question
Applications of integrals
asked 2021-02-13
Find the following integral.
\(\displaystyle\int{\frac{{{3}-{x}}}{{\sqrt{{{x}}}}}}{\left.{d}{x}\right.}\)

Answers (1)

2021-02-14
Given that
\(\displaystyle\int{\frac{{{3}-{x}}}{{\sqrt{{{x}}}}}}{\left.{d}{x}\right.}\)
\(\displaystyle={3}\int{\frac{{{1}}}{{\sqrt{{{x}}}}}}{\left.{d}{x}\right.}-\int\sqrt{{{x}}}{\left.{d}{x}\right.}\)
\(\displaystyle={3}{\left[{\frac{{{x}^{{{\frac{{-{1}}}{{{2}}}}+{1}}}}}{{{\frac{{-{1}}}{{{2}}}}+{1}}}}\right]}-{\left[{\frac{{{x}^{{{\frac{{{1}}}{{{2}}}}+{1}}}}}{{{\frac{{{1}}}{{{2}}}}+{1}}}}\right]}+{C}\)
\(\displaystyle={3}{\left({2}\sqrt{{{x}}}\right)}-{\frac{{{2}}}{{{3}}}}\cdot{x}^{{{\frac{{{3}}}{{{2}}}}}}+{C}\)
\(\displaystyle\therefore\int{\frac{{{3}-{x}}}{{\sqrt{{{x}}}}}}{\left.{d}{x}\right.}={6}\sqrt{{{x}}}-{\frac{{{2}}}{{{3}}}}\cdot{x}^{{{\frac{{{3}}}{{{2}}}}}}+{C}\)
0

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