$\def\hl#1#2{\bbox[#1,1px]{#2}} \def\box#1#2#3#4#5{\color{#2}{\bbox[0px, border: 2px solid #2]{\hl{#3}{\color{white}{\color{#3}{\boxed{\underline{\large\color{#1}{\text{#4}}}\\\color{#1}{#5}\\}}}}}}} \def\verts#1{\left\vert#1\right\vert} \def\Verts#1{\left\Vert#1\right\Vert} \def\pra#1{\left(#1\right)} \def\R{\mathbb{R}}$ $\box{black}{black}{} {Question} {\text{Compute the following line integrals $\int_C\text{F}\cdot d\mathbb{x}$:}\\ \text{(a) F$(x,y)=(x^2,-y)$ and C is the graph of $y=e^x$ from $x=2$ to $x=1$}\\ \text{$(b)$ F$(x,y,z)=(z,-y,x)$ and C is line segment from $(5,0,2)$ to $(5,3,4)$}\\ \text{$(c)$ F$(x,y,z)=(x,y,z^2)$ and C is the intersection of cylinder $x^2+y^2=1$}\\ \text{and $z=x$ oriented counter-clockwise when viewed from above.}}$
My attempts
$(a)$ Consider the parameterization i.e. $(x,y)=\left(1+t,e^{1+t}\right)$, and $(dx,dy)=(1,e^{1+t})$ where $t\in[0,1]$ \begin{align} \int_C\text{F}\cdot d\text{x}=&\int_C x^2dx-ydy\\ =&\int_0^1(1+t)^2-e^{2+2t}dt\\ =&\frac{7}{3} + \frac{e^2-e^4}{2} \end{align}
$(b)$ Let $(x,y,z)=(5,3t,2+2t)$ and $(dx,dy,dz)=(0,3,2)$ where $t\in[0,1]$, have \begin{align} \int_C\text{F}\cdot d\text{x}=&\int_Czdx-ydy+xdz\\ =&\int_0^1-9t+10dt\\ =&\frac{11}{2} \end{align}
$(c)$ Let $(x,y,z)=(\cos(t),\sin(t),\cos(t))$, that $(dx,dy,dz)=(-\sin(t),\cos(t),-\sin(t))$ where $t\in[0,2\pi]$ \begin{align} \int_C\text{F}\cdot d\text{x}=&\int_Cxdx+ydy+z^2dz\\ =&\int_0^{2\pi}-\sin(t)\cos(t)+\sin(t)\cos(t)-\sin(t)\cos^2(t)dt\\ \vdots\\ =&0 \end{align}
Another approach for $(c)$ might be use Stokes' Theorem.
Let $S=\{(x,y,z)\in\R^3:x^2+y^2\le1,z=x\}$, that C is the Stokes' boundary of S. \begin{align} \int_{C}\text{F}\cdot d\text{x}=&\iint_S\nabla\times\text{F}\cdot\text{n}dA\\ =&\iint_S(0,0,0)\cdot\text{n}dA\\ =&0 \end{align} Is my solutions correct?
We can check all your answers in a convenient fashion because all of the vector fields are conservative
a)
$$f(x,y) = \frac{1}{3}x^3-\frac{1}{2}y^2 \implies f(2,e^2)-f(1,e) = \frac{7}{3} - \frac{(e^4-e^2)}{2}$$
b)
$$f(x,y,z) = xz - \frac{1}{2}y^2 \implies f(5,3,4) - f(5,0,2) = \frac{11}{2}$$
c)
$$f(x,y,z) = \frac{x^2+y^2}{2}+\frac{z^3}{3} \implies \int_C \nabla f \cdot dr = 0$$
because it's a closed loop.