Arc Length by Integration

Let $L$ be the length of the curve $\displaystyle{y = \frac{1}{4}{x}^2 - \frac{1}{2} \ln x }$ for $4 \leq x \leq 8.$ Then by the arc length formula, we have $L = \int_{4}^{8} \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\ dx ,$ where $\displaystyle{\frac{dy}{dx} = \frac{1}{2}x-\frac{1}{2x}}.$ Then

$\begin{aligned} L = \int_{4}^{8} \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\ dx &= \int_{4}^{8} \sqrt{1 + { \left(\frac{1}{2}x-\frac{1}{2x} \right)}^2} \ dx \\ &= \int_{4}^{8} \sqrt{{ \left(\frac{1}{2}x+\frac{1}{2x} \right)}^2} \ dx \\ &= \int_{4}^{8} \left ( \frac{1}{2}x + \frac{1}{2x} \right ) \ dx \\ &= \left[ \frac{1}{4}{x}^2 + \frac{1}{2} \ln x \right]_{4}^{8} = \frac{1}{4}( {8}^2 - {4}^2 ) + \frac{1}{2}( \ln 8 - \ln 4 ) \\ &= \frac{48}{4} + \frac{1}{2}\ln \frac{8}{4} = 12 + \frac{1}{2}\ln 2 . \end{aligned}$

Hence, $a + b = 12 + 2 = 14 .$