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Prove:

$$\ \int_C (\sin x - y^2)dx +(x-y \tan^{-1}(y^2))dy = 2.4 $$ where $\ C $ is the curve from $\ (1,2) $ to $\ (-1,2) $ on $\ y = x^2 + 1 $

Using green's theorem

$$\ \int \int_D (Q_x - P_y)dx dy = \int_{-1}^1 \int_{x^2+1}^2 (1+2y)dydx = \frac{28}{5} $$

and that is the intersection of $\ y = 2 $ and $\ y = x^2 +1 $ . Now this area is addition of two line integrals of $\ c_1(t) = (t,2) $ and $\ c_2(t) = (t, t^2+1) $ . so

$$\ \frac{28}{5} = \int_{c_1} F \cdot dr \ + \int_{c_2} F \cdot dr$$

and I'm looking for the value of $\ c_2 $ . but trying to integrate any of them doesn't work. I don't get the correct answer.

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Let path $$C_1: (-t,2) \cup C_2: (t,t^2+1)$$ be a parametrization of $C$ then $$I=\int_{-1}^{1} (-\sin t-4)(-dt)+(-t-2\arctan 4)\times0+ \int_{-1}^{1} (\sin t-(t^2+1)^2)dt+ \left((t-(t^2+1)\arctan(t^2+1)^2)\right)2tdt $$ $$= \left(-2\cos t+4t-\dfrac15t^5-t-\dfrac12\arctan(t^2+1)^2\right)_{-1}^1 = \dfrac{28}{5}$$

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  • $\begingroup$ Thanks but I can not see how does it lead me to the correct answer? $\endgroup$ – bm1125 May 27 at 10:00
  • $\begingroup$ If you mean $2.4$, the correct answer is $\frac{28}{5}$ $\endgroup$ – Nosrati May 27 at 10:06

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