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

Question

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

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Answer

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

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