This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Step 1: Parameterize the line segment C. 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 t 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 t 1. Step 2: Find dx and dy in terms of t. Differentiate x(t) and y(t) with respect to t: (dx)/(dt) = (d)/(dt)(t) = 1 dx = dt (dy)/(dt) = (d)/(dt)(2 + 2t) = 2 dy = 2dt Step 3: Substitute x, y, dx, and dy into the integral. The integral is _C ( y) dy + yx^2 dx. Substitute x=t, y=2+2t, dx=dt, dy=2dt: ( y) dy = ( (2+2t)) (2 dt) Using the periodicity of the sine function, (2 + ) = (): ( (2+2t)) = (2 + 2 t) = (2 t) So, the first term becomes 2 (2 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: _0^1 ( 2 (2 t) dt + (2t^2 + 2t^3) dt ) _0^1 (2t^3 + 2t^2 + 2 (2 t)) dt Step 4: Evaluate the definite integral. Integrate each term with respect to t: (2t^3 + 2t^2 + 2 (2 t)) dt = (2t^4)/(4) + (2t^3)/(3) + 2 ( -((2 t))/(2) ) + C = (t^4)/(2) + (2t^3)/(3) - ((2 t))/() + C Now, evaluate the definite integral from t=0 to t=1: [ (t^4)/(2) + (2t^3)/(3) - ((2 t))/() ]_0^1 = ( ((1)^4)/(2) + (2(1)^3)/(3) - ((2 (1)))/() ) - ( ((0)^4)/(2) + (2(0)^3)/(3) - ((2 (0)))/() ) = ( (1)/(2) + (2)/(3) - ((2))/() ) - ( 0 + 0 - ((0))/() ) Since (2) = 1 and (0) = 1: = ( (1)/(2) + (2)/(3) - (1)/() ) - ( -(1)/() ) = (1)/(2) + (2)/(3) - (1)/() + (1)/() = (1)/(2) + (2)/(3) To add these fractions, find a common denominator, which is 6: = (3)/(6) + (4)/(6) = (7)/(6) The value of the line integral is (7)/(6).