_0^1 x^2+3 x+2 x d x - Vidyadip

Question

$\int_0^1 \frac{x^2+3 x+2}{\sqrt{x}} d x$

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Answer

$
\begin{aligned}
& \text { Let } I =\int_0^1 \frac{x^2+3 x+2}{\sqrt{x}} \cdot d x \\
& =\int_0^1\left(\frac{x^2+3 x+2}{x^{\frac{1}{2}}}\right) \cdot d x \\
& =\int_0^1\left(\frac{x^2}{x^{\frac{1}{2}}}+\frac{3 x}{x^{\frac{1}{2}}}+\frac{2}{x^{\frac{1}{2}}}\right) \cdot d x \\
& =\int_0^1\left(x^{\frac{1}{2}}+3 x^{\frac{1}{2}}+2 x^{\frac{1}{2}}\right) \cdot d x \\
& =\int_0^1 x^{\frac{3}{2}} \cdot d x+3 \int_0^1 x^{\frac{1}{2}} \cdot d x+2 \int_0^1 x^{\frac{1}{2}} \cdot d x \\
& =\left[\frac{\frac{x^5}{2}}{\frac{5}{2}}\right]_0^1+3\left[\frac{\frac{x^3}{2}}{\frac{3}{2}}\right]_0^1+2\left[\frac{x^1}{\frac{1}{2}}\right]_0^1 \\
& =\frac{2}{5}(1-0) 3 \times \frac{2}{3}(1-0)+2 \times 2(1-0) \\
& =\frac{2}{5} 2+4 \\
& \therefore I =\frac{32}{5} .
\end{aligned}
$

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