Evaluate the line integral integral_C (pi y) dy + yx2 dx, where C is the line segment from (0,2) to (1,4).

The line segment goes from $(x_0, y_0) = (0,2)$ to $(x_1, y_1) = (1,4)$.
A parameterization for a line segment is given by:

$$x(t) = x_0 + (x_1 - x_0)t$$ $$y(t) = y_0 + (y_1 - y_0)t$$

for $0 \le t \le 1$.

Substitute the given points:

$$x(t) = 0 + (1 - 0)t = t$$ $$y(t) = 2 + (4 - 2)t = 2 + 2t$$

So, $x = t$ and $y = 2 + 2t$ for $0 \le t \le 1$.

Differentiate $x(t)$ and $y(t)$ with respect to $t$:

$$\frac{dx}{dt} = \frac{d}{dt}(t) = 1 \implies dx = dt$$ $$\frac{dy}{dt} = \frac{d}{dt}(2 + 2t) = 2 \implies dy = 2dt$$

The integral is $\int_C \sin(\pi y) dy + yx^2 dx$.
Substitute $x=t$, $y=2+2t$, $dx=dt$, $dy=2dt$:

$$\sin(\pi y) dy = \sin(\pi (2+2t)) (2 dt)$$

Using the periodicity of the sine function, $\sin(2\pi + \theta) = \sin(\theta)$:

$$\sin(\pi (2+2t)) = \sin(2\pi + 2\pi t) = \sin(2\pi t)$$

So, the first term becomes $2 \sin(2\pi t) dt$.

The second term is $yx^2 dx$:

$$yx^2 dx = (2+2t)(t^2) dt = (2t^2 + 2t^3) dt$$

Now, substitute these into the integral, changing the limits of integration from $t=0$ to $t=1$:

$$\int_0^1 \left( 2 \sin(2\pi t) dt + (2t^2 + 2t^3) dt \right)$$ $$\int_0^1 (2t^3 + 2t^2 + 2 \sin(2\pi t)) dt$$

Integrate each term with respect to $t$:

$$\int (2t^3 + 2t^2 + 2 \sin(2\pi t)) dt = \frac{2t^4}{4} + \frac{2t^3}{3} + 2 \left( -\frac{\cos(2\pi t)}{2\pi} \right) + C$$ $$= \frac{t^4}{2} + \frac{2t^3}{3} - \frac{\cos(2\pi t)}{\pi} + C$$

Now, evaluate the definite integral from $t=0$ to $t=1$:

$$\left[ \frac{t^4}{2} + \frac{2t^3}{3} - \frac{\cos(2\pi t)}{\pi} \right]_0^1$$ $$= \left( \frac{(1)^4}{2} + \frac{2(1)^3}{3} - \frac{\cos(2\pi (1))}{\pi} \right) - \left( \frac{(0)^4}{2} + \frac{2(0)^3}{3} - \frac{\cos(2\pi (0))}{\pi} \right)$$ $$= \left( \frac{1}{2} + \frac{2}{3} - \frac{\cos(2\pi)}{\pi} \right) - \left( 0 + 0 - \frac{\cos(0)}{\pi} \right)$$

Since $\cos(2\pi) = 1$ and $\cos(0) = 1$:

$$= \left( \frac{1}{2} + \frac{2}{3} - \frac{1}{\pi} \right) - \left( -\frac{1}{\pi} \right)$$ $$= \frac{1}{2} + \frac{2}{3} - \frac{1}{\pi} + \frac{1}{\pi}$$ $$= \frac{1}{2} + \frac{2}{3}$$

To add these fractions, find a common denominator, which is 6:

$$= \frac{3}{6} + \frac{4}{6}$$ $$= \frac{7}{6}$$

The value of the line integral is $\frac{7}{6}$.