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You need to find the parametrization of the curve:
`x=(1-t)x_0 + tx_1=gt x = (1-t)(-1) + 2t=gt x = -1 + t + 2t=gtx=3t-1`
`y=(1-t)y_0 + ty_1=gty = (1-t)(1) + 4t=gty=3t+1`
The individual parametric equations are x=3t-1 and y=3t+1.
You need to determine `ds = sqrt((dx/dx)^2 + (dy/dt)^2)dt`
`dx/dt = (3t-1)' =gt dx/dt = 3`
`` `dy/dt = (3t+1)' =gt dx/dt = 3`
`ds = sqrt(3^2 + 3^2)dt =gt ds = sqrt9 dt`
Write the function f(x,y) in terms of t:
`f(x,y) = 2(3t-1)(3t+1) = 2(9t^2 - 1)`
Calculating the line integral yields:
`oint` f(x,y)ds = `2sqrt9 int_0^1 (9t^2 - 1)dt = 2sqrt9(9t^3/3 - t)(0-gt1)`
`` `oint f(x,y)ds = 2sqrt9 (3*1^3 - 3*0^3 - 1 + 0) = 4sqrt9`
Evaluating the line integral yields `oint f(x,y)ds = 4sqrt9` .
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