Question Let the curve \( C \) be the function \( f(y)=6 y^{5}+1 \) over the interval \( 1 \leq y \leq 7 \). Set up the integral for the arc length of the curve \( C \).

To find the arc length of the curve \( C \) defined by \( f(y) = 6y^5 + 1 \) over the interval \( 1 \leq y \leq 7 \), we can use the arc length formula for a function expressed in terms of \( y \): \[ \text{Arc Length} = \int_{a}^{b} \sqrt{1 + \left( \frac{df}{dy} \right)^2} \, dy \] **Step-by-Step Setup:** 1. **Identify the function and its derivative:** Given \( f(y) = 6y^5 + 1 \), compute the derivative with respect to \( y \): \[ \frac{df}{dy} = \frac{d}{dy}(6y^5 + 1) = 30y^4 \] 2. **Substitute into the arc length formula:** Plug \( \frac{df}{dy} = 30y^4 \) into the formula: \[ \text{Arc Length} = \int_{1}^{7} \sqrt{1 + (30y^4)^2} \, dy \] 3. **Simplify the integrand:** \[ \sqrt{1 + (30y^4)^2} = \sqrt{1 + 900y^8} \] 4. **Finalize the integral expression:** \[ \text{Arc Length} = \int_{1}^{7} \sqrt{1 + 900y^8} \, dy \] **Final Integral:** \[ \boxed{\displaystyle \int_{1}^{7} \sqrt{\,1\,+\,900\,y^{8}\,} \;\mathrm{d}y} \]