9-24 Find the exact length of the curve. 9. \( y=\frac{2}{3} x^{3 / 2}, \quad 0 \leqslant x \leqslant 2 \)

To find the exact length of the curve given by \( y = \frac{2}{3}x^{3/2} \) over the interval \( 0 \leqslant x \leqslant 2 \), we can use the formula for the arc length of a curve: \[ L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \] First, we need to find \( \frac{dy}{dx} \) for the given function: \[ y = \frac{2}{3}x^{3/2} \] Taking the derivative with respect to \( x \): \[ \frac{dy}{dx} = \frac{2}{3} \cdot \frac{3}{2}x^{1/2} = x^{1/2} \] Now, we can substitute \( \frac{dy}{dx} \) into the formula for the arc length: \[ L = \int_{0}^{2} \sqrt{1 + (x^{1/2})^2} \, dx \] \[ L = \int_{0}^{2} \sqrt{1 + x} \, dx \] Next, we can calculate the integral to find the exact length of the curve. Evaluate the integral by following steps: - step0: Evaluate using substitution: \(\int_{0}^{2} \sqrt{1+x} dx\) - step1: Evaluate the power: \(\int_{0}^{2} \left(1+x\right)^{\frac{1}{2}} dx\) - step2: Evaluate the integral: \(\int \left(1+x\right)^{\frac{1}{2}} dx\) - step3: Use the substitution \(t=1+x\) to transform the integral\(:\) \(\int t^{\frac{1}{2}} dt\) - step4: Evaluate the integral: \(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\) - step5: Simplify: \(\frac{2}{3}t^{\frac{3}{2}}\) - step6: Substitute back: \(\frac{2}{3}\left(1+x\right)^{\frac{3}{2}}\) - step7: Return the limits: \(\left(\frac{2}{3}\left(1+x\right)^{\frac{3}{2}}\right)\bigg |_{0}^{2}\) - step8: Calculate the value: \(2\times 3^{\frac{1}{2}}-\frac{2}{3}\) - step9: Transform the expression: \(2\sqrt{3}-\frac{2}{3}\) - step10: Calculate: \(\frac{6\sqrt{3}-2}{3}\) The exact length of the curve given by \( y = \frac{2}{3}x^{3/2} \) over the interval \( 0 \leqslant x \leqslant 2 \) is \( \frac{6\sqrt{3}-2}{3} \).