# Using Green's theorem to compute integral on curve

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.

Let path $$C_1: (-t,2) \cup C_2: (t,t^2+1)$$ be a parametrization of $$C$$ then $$I=\int_{-1}^{1} (-\sin t-4)(-dt)+(-t-2\arctan 4)\times0+ \int_{-1}^{1} (\sin t-(t^2+1)^2)dt+ \left((t-(t^2+1)\arctan(t^2+1)^2)\right)2tdt$$ $$= \left(-2\cos t+4t-\dfrac15t^5-t-\dfrac12\arctan(t^2+1)^2\right)_{-1}^1 = \dfrac{28}{5}$$
• If you mean $2.4$, the correct answer is $\frac{28}{5}$ – Nosrati May 27 at 10:06