Answer
$\displaystyle \frac{2}{3}$
Work Step by Step
FTC: Let $f$ be a continuous function defined on the interval $[a, b]$,
and let $F$ be any antiderivative of f defined on $[a, b]$. Then
$\displaystyle \int_{a}^{b}f(x)dx=\left[F(x)\right]_{a}^{b}=F(b)-F(a)$.
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$f(x)=x^{1/2}$, which is continuous on [$0,1$].
First, find an antiderivative $F(x)$:
$\displaystyle \int f(x)dx=\frac{x^{3/2}}{\frac{3}{2}}+C=\frac{2x^{3/2}}{3}+C$
(we take C=0, since by FTC we need ANY antiderivative)
$F(x)=\displaystyle \frac{2x^{3/2}}{3}$
Next, evaluate:
$F(1)=\displaystyle \frac{2}{3}$
$F(0)=0$
Finally, apply the theorem:
$\displaystyle \int_{0}^{1}f(x)dx=F(1)-F(0)=\frac{2}{3}-0$
$=\displaystyle \frac{2}{3}$
