Let $C$ represent the parabolic path $y = x^2 - 1$ from the $(0,-1)$ to $(1,0)$. Using the fundamental theorem of line integrals, evaluate
$$ \int_C(ye^x-\sin x)dx+(e^x+2)dy $$
I know that the fundamental theorem of line integrals states that
$$ \int_C\boldsymbol{\nabla} f \cdot dr=f(r(b))-f(r(a)) $$
but how do you apply it to this problem?
You want to find a scalar function $f$ such that $$\nabla f = (ye^x-\sin x, e^x+2).$$ For that, we can take $$f(x,y) = ye^x + \cos x + 2y.$$ Then, $$\int_C (ye^x - \sin x) \ dx + (e^x + 2) \ dy = \int_C \nabla f \cdot dr = f(1,0) - f(0,-1) = \cos 1 + 2.$$
You can check this by just parametrizing the curve ($r(t) = (t,t^2-1)$ for $0\leq t\leq1$) and computing the integral: $$\int_C (ye^x - \sin x) \ dx + (e^x + 2) \ dy = \int_0^1 (t^2-1)e^t - \sin t + 2t(e^t + 2) \ dt.$$ This is obviously more involved, but a few applications of integration by parts should take care of non-trivial parts.