MHB How to Compute the Line Integral Over a Piecewise Curve?

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To compute the line integral over the piecewise curve, first define the two segments of the curve: C1 for 0≤t≤1 with x(t)=t and y(t)=t^2, and C2 for 1≤t≤2 with x(t)=2-t and y(t)=2-t. The integral can be expressed as the sum of the integrals over these two segments, ∫C = ∫C1 + ∫C2. Substitute the parameterizations into the integral expression ∫C {(-x^2 + y^2)dx + xydy} for each segment. Finally, evaluate the integrals for both segments to find the total value of the line integral.
aruwin
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I have no idea how to even start with this problem. I know the basics but this one just gets complicated. Please guide me!Find the line integral:
∫C {(-x^2 + y^2)dx + xydy}
When 0≤t≤1 for the curved line C, x(t)=t, y(t)=t^2
and when 1≤t≤2, x(t)= 2 - t , y(t) = 2-t.
Use x(t) and y(t) and C={(x(t),y(t))|0≤t≤2}
Help!
 
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It looks to me as though you could define

$$C_{1}:\quad 0\le t\le 1,\quad x=t,\quad y=t^{2},$$
and
$$C_{2}:\quad 1\le t\le 2,\quad x=2-t,\quad y=2-t.$$

You're asked to compute
$$\int_{C}=\int_{C_{1}}+\int_{C_{2}}.$$
Where do you go from here?
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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