Answer
\[I = 1-\pi\]
Work Step by Step
Step 1: In our case, we have the curve: \[ x=2\cos t, \quad y=4\sin t, \quad 0\leq t\leq \frac{\pi}{4} \] Therefore: \[ I = \int_C (x+2y)dx + (x-y)dy = \int_0^{\pi/4} [(2\cos t+8\sin t)(-2\sin t)+(2\cos t-4\sin t)4\cos t]dt \] \[ = \int_0^{\pi/4} (-4\sin t\cos t-16\sin^2 t+8\cos^2 t-16\sin t\cos t)dt \] \[ = \int_0^{\pi/4} (-2\sin^2 t-8\sin^2 t+8\cos^2 t-8\sin^2 t)dt \] \[ = \int_0^{\pi/4} [-8(1+\cos^2 t)+8\cos^2 t-10\sin^2 t]dt \] \[ = \int_0^{\pi/4} [-8\left(1+\frac{\cos^2 t}{2}\right)+8\cos^2 t-10\sin^2 t]dt \] \[ = \left[-4(t-2\sin^2 t)+4\sin^2 t+5\cos^2 t\right]_0^{\pi/4} \] \[ = [-4(\pi/4-1/2)+1/2+5(1)-0] \] \[ = [-(\pi/2-1/2)+1/2+5] \] \[ = [1-\pi] \] Result: \[ I = 1-\pi \]
