Evaluate _0^1 (x^2-x ) x d x. - Vidyadip

Question

Evaluate $\int_0^1 \frac{\left(x^2-x\right)}{\sqrt{x}} d x$.

✓

Answer

Let
$
\begin{aligned}
I & =\int_0^1 \frac{\left(x^2-x\right)}{\sqrt{x}} d x=\int_0^1\left(x^{3 / 2}-x^{\frac{1}{2}}\right) d x \\
I & =\left[\frac{x^{5 / 2}}{5 / 2}-\frac{x^{3 / 2}}{3 / 2}\right]_0^1=\left[\frac{2}{5}(1-0)-\frac{2}{3}(1-0)\right] \\
& =\frac{2}{5}-\frac{2}{3}=\frac{6-10}{15}=\frac{-4}{15} \text { }
\end{aligned}
$

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