Answer
$$\frac 8 {15} (\sqrt 2 + 1)$$
Work Step by Step
1. According to equation 3:
$$\int_C f(x,y)ds = \int_a^bf(x(t),y(t))\sqrt {(\frac {dx}{dt})^2 + (\frac{dy}{dt})^2}dt$$
2. Find dx/dy, dy/dt and f(x(t),y(t)):
$$\frac{dx}{dt} = \frac{d(t^2)}{dt} = 2t$$ $$\frac{dy}{dt} = \frac{d(2t)}{dt} = 2$$ $$f(x(t),y(t)) = xy = t^2*2t = 2t^3$$
3. Integrate:
$$\int_0^1 (2t^3) \sqrt {(2t)^2 + (2)^2}dt = \int_0^1 (2t^3) \sqrt {4t^2 + 4}dt$$ $$\int_0^1 (2t^3) \sqrt {4(t^2 + 1)}dt = \int_0^1 (2t^3)(2) \sqrt {t^2 + 1}dt$$
$u = t^2 + 1 \longrightarrow t^2 = u - 1$
$u_1 = 0^2 + 1= 1$
$u_2 = 1^2 + 1 = 2$
$du = 2t \space dt$
$$ \int_1^2 (2t^3)(2) \sqrt {u} \space dt = 2 \int_1^2 (t^2) \sqrt u \space 2t \space dt$$ $$ 2 \int_1^2 (u-1) \sqrt u \space du = 2 \int_1^2 u^{3/2} - u^{1/2} \space du$$
$$2[\frac 25 u^{5/2} - \frac 23 u^{3/2}]_1^2$$ $$ 2(\frac 25 2^{5/2} - \frac 23 2^{3/2}) - 2(\frac 25 1^{5/2} - \frac 23 1^{3/2})$$ $$2(\frac 85 \sqrt 2 - \frac 43 \sqrt 2) - 2 (\frac 25 - \frac 23)$$ $$2(\frac{24}{15}\sqrt 2 - \frac {20} {15} \sqrt 2) - 2(\frac 6 {15} - \frac {10}{15})$$ $$\frac{8}{15}\sqrt 2 + \frac 8 {15}$$ $$\frac 8 {15} (\sqrt 2 + 1)$$
