math

Use the method explained in class to find the pat integral: c f*dc

where f=<2x-4y^2, 2y-3x^2> c(t)=<t^2+t,t^3>

from t=-1 to t =2

$ Solution:
\int_C F \cdot dc
\bg_white F = \left \langle 2x-4y^2 \ , \ 2y-3x^2 \right \rangle
\bg_white C(t) = \left \langle t^2 + t \ , \ t^3 \right \rangle
\bg_white \Rightarrow \ \ \ C'(t) = \left \langle 2t+1 \ , \ 3t^2 \right \rangle
F(C(t)) = \left \langle 2(t^2+t)-4(t^3)^2 \ , \ 2(t^3)-3(t^2+t)^2 \right \rangle
F(C(t)) = \left \langle -4 t^6 + 2 t^2 + 2 t \ , \ -3 t^4 - 4 t^3 - 3 t^2 \right \rangle
\\
\int_C F \cdot dc = \int_{t = -1 }^{2} F(C(t)) \cdot C'(t) \ dt
          = \int_{t = -1 }^{2} \left \langle -4 t^6 + 2 t^2 + 2 t \ , \ -3 t^4 - 4 t^3 - 3 t^2 \right \rangle \cdot \left \langle 2t+1 \ , \ 3t^2 \right \rangle \ dt

          = \int_{t = -1 }^{2} ((-4 t^6 + 2 t^2 + 2 t)(2t+1) \ + \ (-3 t^4 - 4 t^3 - 3 t^2)(3t^2)) \ dt

          = \int_{t = -1 }^{2} (-8 t^7 - 13 t^6 - 12 t^5 - 9 t^4 + 4 t^3 + 6 t^2 + 2 t) \ dt

          =\left [ - t^8 - \frac{13 t^7}{7} - 2 t^6 - \frac{9 t^5}{5} + t^4 + 2 t^3 + t^2 \right ]_{t = -1 }^{2}

          =\left [ - (2)^8 - \frac{13 (2)^7}{7} - 2 (2)^6 - \frac{9 (2)^5}{5} + (2)^4 + 2 (2)^3 + (2)^2 \right ] - \left [ - (-1)^8 - \frac{13 (-1)^7}{7} - 2 (-1)^6 - \frac{9 (-1)^5}{5} + (-1)^4 + 2 (-1)^3 + (-1)^2 \right ]

=-\frac{22516}{35} - \left [ \frac{23}{35} \right ]

=-\frac{22516}{35} - \frac{23}{35}

=-\frac{22539}{35}