The length of the curve \[{\text{y}} = \frac{2}{3}{{\text{x}}^{\frac{3}{2}}}\] between x = 0 and x = 1 is A. 0.27 B. 0.67 C. 1 D. 1.22

0.27
0.67
1
1.22

The correct answer is $\boxed{\text{B) }0.67}$.

The formula for the arc length of a curve $y=f(x)$ over the interval $[a,b]$ is:

$$L = \int_a^b \sqrt{1+\left(\frac{dy}{dx}\right)^2} dx$$

In this case, we have $y=\frac{2}{3}x^{3/2}$ and $a=0$ and $b=1$. Substituting these values into the formula, we get:

$$L = \int_0^1 \sqrt{1+\left(\frac{d}{dx}\left(\frac{2}{3}x^{3/2}\right)\right)^2} dx = \int_0^1 \sqrt{1+\left(\frac{2}{3}\cdot\frac{3}{2}x^{^{-1/2}}\right)^2} dx = \int_0^1 \sqrt{1+\frac{4}{9}x^{-1}} dx$$

We can now evaluate this integral using the substitution $u=\frac{1}{x}$. This gives us:

$$L = \int_0^1 \sqrt{1+\frac{4}{9}x^{-1}} dx = \int_1^\infty \sqrt{1+\frac{4}{9}u^2} du = \left.\frac{2}{3}u^{3/2}\sqrt{1+\frac{4}{9}u^2}\right|_1^\infty = \frac{2}{3}\sqrt{1+\frac{4}{9}} – \frac{2}{3} = 0.67$$

Therefore, the length of the curve $y=\frac{2}{3}x^{3/2}$ between $x=0$ and $x=1$ is $\boxed{\text{B) }0.67}$.

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