Question

Find the following integral. \int \sqrt{x}(x^{3}+\frac{x}{2})dx

Applications of integrals
ANSWERED
asked 2021-01-13
Find the following integral.
\(\displaystyle\int\sqrt{{{x}}}{\left({x}^{{{3}}}+{\frac{{{x}}}{{{2}}}}\right)}{\left.{d}{x}\right.}\)

Answers (1)

2021-01-14
Given that
\(\displaystyle\int\sqrt{{{x}}}{\left({x}^{{{3}}}+{\frac{{{x}}}{{{2}}}}\right)}{\left.{d}{x}\right.}\)
\(\displaystyle=\int{\left({x}^{{{3}+{\frac{{{1}}}{{{2}}}}}}+{\frac{{{x}^{{{\frac{{{1}}}{{{2}}}}\cdot{x}}}}}{{{2}}}}\right)}{\left.{d}{x}\right.}\)
\(\displaystyle=\int{\left({x}^{{{\frac{{{7}}}{{{2}}}}}}+{\frac{{{x}^{{{\frac{{{3}}}{{{2}}}}}}}}{{{2}}}}\right)}{\left.{d}{x}\right.}\)
\(\displaystyle={\frac{{{x}^{{{\frac{{{7}}}{{{2}}}}+{1}}}}}{{{\frac{{{7}}}{{{2}}}}+{1}}}}+{\frac{{{x}^{{{\frac{{{3}}}{{{2}}}}+{1}}}}}{{{\frac{{{\frac{{{3}}}{{{2}}}}+{1}}}{{{2}}}}}}}+{C}\)
\(\displaystyle={\frac{{{2}}}{{{9}}}}{x}^{{{\frac{{{9}}}{{{2}}}}}}+{\frac{{{1}}}{{{2}}}}\cdot{\frac{{{2}}}{{{5}}}}\cdot{x}^{{{\frac{{{5}}}{{{2}}}}}}+{C}\)
\(\displaystyle={\frac{{{2}}}{{{9}}}}{x}^{{{\frac{{{9}}}{{{2}}}}}}+{\frac{{{1}}}{{{5}}}}\cdot{x}^{{{\frac{{{5}}}{{{2}}}}}}+{C}\)
\(\displaystyle\therefore\int\sqrt{{{x}}}{\left({x}^{{{3}}}+{\frac{{{x}}}{{{2}}}}\right)}{\left.{d}{x}\right.}={\frac{{{2}}}{{{9}}}}\cdot{x}^{{{\frac{{{9}}}{{{2}}}}}}+{\frac{{{1}}}{{{5}}}}\cdot{x}^{{{\frac{{{5}}}{{{2}}}}}}+{C}\)
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