Calculate $\int_\gamma fdx$ for $f(x)=(x_1^2+5x_2+3x_2x_3,5x_1+3x_1x_3-2,3x_1x_2-4x_3)$ and $\gamma$ is the helix with radius $1$ and pitch $1$ from $(0, 1, 0)$ to $(0, 1, 2\pi)$
Now, the coordinates of a circular helix of radius $a$ and pitch $2\pi b$ are: $(x(t),y(t),z(t))=(a\cos{t},a\sin{t},bt)=(\cos{t},\sin{t},t)$ since $a = 1$ and $b=1$
My definition for calculating the line integral would be: $\int_\gamma fdx=\int(f(\gamma(t)))·\gamma'$
My doubt here is the parametrization of $\gamma(t)$, I don't know for which values of $t$ :
Let $\gamma(t)=(\cos{t},\sin{t},t)$, for $t\in[0,\pi]$? For $t\in[0,2\pi]$? For $t\in$ something else?
As $\gamma$ starts at $(0,1,0)$, and not at $(1,0,0)$, you have $$\gamma:\quad t\mapsto \gamma(t)=(-\sin t,\cos t, t)\qquad(0\leq t\leq2\pi)\ ,$$ and then have to calculate $$\int_\gamma f\cdot dx=\int_0^{2\pi}f\bigl(\gamma(t)\bigr)\cdot\gamma'(t)\>dt\ .$$ This example has a special flavor: The force field $f$ is in fact the gradient of some scalar function $F$, i.e., $$f(x_1,x_2,x_3)=\nabla F(x_1,x_2,x_3)\ .$$ If you can guess $F$ the value of the integral is then given by $$\int_\gamma f\cdot dx=F(0,1,2\pi)-F(0,1,0)\ .$$ In fact $F(x_1,x_2,x_3)={1\over3}x_1^3+5x_1x_2+3x_1x_2x_3-2x_2-2x_3^2$.