1. **Find the arc length of** $y = \frac{x^4}{8} + \frac{1}{4x^2}$ from $x=1$ to $x=2$. 2. **Find the arc length of** $(y-1)^3 = x^2$ on the interval $[0,8]$. 3. **Find the arc length of** $y = 4 - x^2$ on $0 \leq x \leq 2$. --- ### Step 1: Recall the arc length formula for $y=f(x)$ The arc length $L$ from $x=a$ to $x=b$ is given by: $$ L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx $$ ### Problem 1: $y = \frac{x^4}{8} + \frac{1}{4x^2}$, $x \in [1,2]$ 1. Compute $\frac{dy}{dx}$: $$ \frac{dy}{dx} = \frac{d}{dx}\left(\frac{x^4}{8} + \frac{1}{4x^2}\right) = \frac{4x^3}{8} - \frac{2}{4x^3} = \frac{x^3}{2} - \frac{1}{2x^3} $$ 2. Square the derivative: $$ \left(\frac{dy}{dx}\right)^2 = \left(\frac{x^3}{2} - \frac{1}{2x^3}\right)^2 = \frac{1}{4}\left(x^3 - \frac{1}{x^3}\right)^2 = \frac{1}{4}\left(x^6 - 2 + \frac{1}{x^6}\right) = \frac{x^6}{4} - \frac{1}{2} + \frac{1}{4x^6} $$ 3. Add 1 inside the square root: $$ 1 + \left(\frac{dy}{dx}\right)^2 = 1 + \frac{x^6}{4} - \frac{1}{2} + \frac{1}{4x^6} = \frac{1}{2} + \frac{x^6}{4} + \frac{1}{4x^6} $$ 4. Rewrite as: $$ \frac{1}{2} + \frac{x^6}{4} + \frac{1}{4x^6} = \frac{2}{4} + \frac{x^6}{4} + \frac{1}{4x^6} = \frac{x^6 + 2 + \frac{1}{x^6}}{4} = \frac{\left(x^3 + \frac{1}{x^3}\right)^2}{4} $$ 5. So the integrand simplifies to: $$ \sqrt{1 + \left(\frac{dy}{dx}\right)^2} = \sqrt{\frac{\left(x^3 + \frac{1}{x^3}\right)^2}{4}} = \frac{1}{2} \left| x^3 + \frac{1}{x^3} \right| $$ Since $x \geq 1$, the expression inside is positive, so: $$ = \frac{1}{2} \left(x^3 + \frac{1}{x^3}\right) $$ 6. The arc length integral is: $$ L = \int_1^2 \frac{1}{2} \left(x^3 + \frac{1}{x^3}\right) dx = \frac{1}{2} \int_1^2 \left(x^3 + x^{-3}\right) dx $$ 7. Integrate term by term: $$ \int x^3 dx = \frac{x^4}{4}, \quad \int x^{-3} dx = \frac{x^{-2}}{-2} = -\frac{1}{2x^2} $$ 8. Evaluate: $$ L = \frac{1}{2} \left[ \frac{x^4}{4} - \frac{1}{2x^2} \right]_1^2 = \frac{1}{2} \left( \left(\frac{2^4}{4} - \frac{1}{2 \cdot 2^2}\right) - \left(\frac{1^4}{4} - \frac{1}{2 \cdot 1^2}\right) \right) $$ 9. Calculate values: $$ \frac{2^4}{4} = \frac{16}{4} = 4, \quad \frac{1}{2 \cdot 4} = \frac{1}{8} $$ $$ \frac{1}{4} = 0.25, \quad \frac{1}{2} = 0.5 $$ 10. Substitute: $$ L = \frac{1}{2} \left(4 - \frac{1}{8} - 0.25 + 0.5\right) = \frac{1}{2} \left(4 - 0.125 + 0.25\right) = \frac{1}{2} \times 4.125 = 2.0625 $$ --- ### Problem 2: $(y-1)^3 = x^2$, $x \in [0,8]$ 1. Express $y$ in terms of $x$: $$ (y-1)^3 = x^2 \implies y-1 = x^{2/3} \implies y = 1 + x^{2/3} $$ 2. Compute $\frac{dy}{dx}$: $$ \frac{dy}{dx} = \frac{2}{3} x^{-1/3} $$ 3. Square the derivative: $$ \left(\frac{dy}{dx}\right)^2 = \frac{4}{9} x^{-2/3} $$ 4. Arc length formula: $$ L = \int_0^8 \sqrt{1 + \frac{4}{9} x^{-2/3}} \, dx = \int_0^8 \sqrt{1 + \frac{4}{9} \frac{1}{x^{2/3}}} \, dx $$ 5. Rewrite inside the root: $$ = \int_0^8 \sqrt{1 + \frac{4}{9} x^{-2/3}} \, dx $$ 6. Substitute $u = x^{1/3}$, so $x = u^3$, $dx = 3u^2 du$. 7. Change limits: when $x=0$, $u=0$; when $x=8$, $u=2$. 8. Rewrite integral: $$ L = \int_0^2 \sqrt{1 + \frac{4}{9} u^{-2}} \cdot 3u^2 du = 3 \int_0^2 u^2 \sqrt{1 + \frac{4}{9u^2}} du $$ 9. Simplify inside the root: $$ 1 + \frac{4}{9u^2} = \frac{9u^2 + 4}{9u^2} $$ 10. So integrand: $$ 3 u^2 \sqrt{\frac{9u^2 + 4}{9u^2}} = 3 u^2 \frac{\sqrt{9u^2 + 4}}{3u} = u \sqrt{9u^2 + 4} $$ 11. Integral becomes: $$ L = \int_0^2 u \sqrt{9u^2 + 4} \, du $$ 12. Use substitution $w = 9u^2 + 4$, $dw = 18u du$, so $u du = \frac{dw}{18}$. 13. Rewrite integral: $$ L = \int \sqrt{w} \frac{dw}{18} = \frac{1}{18} \int w^{1/2} dw = \frac{1}{18} \cdot \frac{2}{3} w^{3/2} + C = \frac{1}{27} w^{3/2} + C $$ 14. Evaluate from $u=0$ to $u=2$: $$ w = 9u^2 + 4 $$ At $u=2$, $w=9 \cdot 4 + 4 = 40$. At $u=0$, $w=4$. 15. So: $$ L = \frac{1}{27} \left(40^{3/2} - 4^{3/2}\right) = \frac{1}{27} \left(40 \sqrt{40} - 8\right) $$ --- ### Problem 3: $y = 4 - x^2$, $x \in [0,2]$ 1. Compute $\frac{dy}{dx}$: $$ \frac{dy}{dx} = -2x $$ 2. Square the derivative: $$ \left(\frac{dy}{dx}\right)^2 = 4x^2 $$ 3. Arc length formula: $$ L = \int_0^2 \sqrt{1 + 4x^2} \, dx $$ 4. Use substitution for integral: $$ \int \sqrt{1 + 4x^2} \, dx $$ 5. Use formula for $\int \sqrt{a^2 + u^2} du$: $$ \int \sqrt{1 + 4x^2} dx = \frac{x}{2} \sqrt{1 + 4x^2} + \frac{\sinh^{-1}(2x)}{4} + C $$ 6. Evaluate from 0 to 2: $$ L = \left[ \frac{x}{2} \sqrt{1 + 4x^2} + \frac{\sinh^{-1}(2x)}{4} \right]_0^2 $$ 7. Calculate values: $$ \frac{2}{2} \sqrt{1 + 16} = 1 \times \sqrt{17} = \sqrt{17} $$ $$ \sinh^{-1}(4) = \ln(4 + \sqrt{17}) $$ 8. So: $$ L = \sqrt{17} + \frac{1}{4} \ln(4 + \sqrt{17}) $$