A recycled curve

We require that $\displaystyle \int y dx = \int \sqrt{1 + (y')^{2}} dx$ over any interval. As the interval is arbitrary, the integrands must be equal. Thus

$y = \sqrt{1 + (y')^{2}} \Longrightarrow y^{2} = 1 + \left(\dfrac{dy}{dx}\right)^{2} \Longrightarrow \sqrt{y^{2} - 1} = \dfrac{dy}{dx} \Longrightarrow \dfrac{dy}{\sqrt{y^{2} - 1}} = dx \Longrightarrow$

$\cosh^{-1}(y) = x + C \Longrightarrow y = \cosh(x + C)$ .

Now to have $f(0) = 1$ we must have $\cosh(C) = 1 \Longrightarrow C = 0$ , so $y = f(x) = \cosh(x)$ . Since the integral of $\cosh(x)$ is $\sinh(x)$ , the given integral is equal to

$\sinh(\ln(5)) - \sinh(\ln(2)) = \dfrac{5 - \dfrac{1}{5}}{2} - \dfrac{2 - \dfrac{1}{2}}{2} = \dfrac{12}{5} - \dfrac{3}{4} = \dfrac{33}{20}$ , and so $a + b + 33 + 20 = \boxed{53}$ .

Note that $\sinh(x) = \dfrac{e^{x} - e^{-x}}{2}$ .

Knowing the area between the curve and the x axis is equal to the arc length, we can derive the equation $\int_{a}^{b}f(x)dx = \int_{a}^{b}\sqrt{1+(f'(x))^2}dx$ Taking the derivative of both sides gets us $f(x) = \sqrt{1+(f'(x))^2} \\ (f(x))^2 = 1+(f'(x))^2 \\ (f(x))^2-(f'(x))^2 = 1$ Recalling that $\frac{d}{dx} \cosh(x) = \sinh(x)$ and $\cosh(x)-\sinh(x) = 1$ , we know $f(x) = \cosh(x+C)$ . Knowing $f(0) = 1$ and $\cosh(0) = 1$ , we know $C = 0$ . Now we can solve for the integral $\int_{\ln(2)}^{\ln(5)}\cosh(x)dx \\ [\sinh(x)]_{\ln(2)}^{\ln(5)} \\ \frac{e^{\ln(5)}-e^{-\ln(5)}}{2} - \frac{e^{\ln(5)}-e^{-\ln(5)}}{2} \\ \frac{12}{5}-\frac{3}{4} \\ \frac{48-15}{20} = \frac{33}{20}$ Therefore, $a = 33$ and $b=20$ . Thus, $a+b = 33+20 = \boxed{53}$