It is not necessary to calculate an integral.

An exact value for the integral can be determined, including instances when evaluating the integral without the Fundamental Theorem for Line Integrals would require an approximation technique because an antiderivative cannot be found for the integrand.

Both of the above

None of the above

${\int <!--\; \int \; -->}_{C_1}^{}\stackrel{<!--\; \rightharpoonup \; -->}{F}\cdot d\stackrel{<!--\; \rightharpoonup \; -->}{r}$
$\le {\int <!--\; \int \; -->}_{C_2}^{}\stackrel{}{F}\cdot d\stackrel{<!--\; \rightharpoonup \; -->}{r}$
$\le {\int <!--\; \int \; -->}_{C_3}^{}\stackrel{<!--\; \rightharpoonup \; -->}{F}\cdot d\stackrel{<!--\; \rightharpoonup \; -->}{r}$
$\le {\int <!--\; \int \; -->}_{C_4}^{}\stackrel{<!--\; \rightharpoonup \; -->}{F}\cdot d\stackrel{<!--\; \rightharpoonup \; -->}{r}$

${\int <!--\; \int \; -->}_{C_4}^{}\stackrel{<!--\; \rightharpoonup \; -->}{F}\cdot d\stackrel{<!--\; \rightharpoonup \; -->}{r}$
$\le {\int <!--\; \int \; -->}_{C_2}^{}\stackrel{}{F}\cdot d\stackrel{<!--\; \rightharpoonup \; -->}{r}$
$\le {\int <!--\; \int \; -->}_{C_3}^{}\stackrel{<!--\; \rightharpoonup \; -->}{F}\cdot d\stackrel{<!--\; \rightharpoonup \; -->}{r}$
$\le {\int <!--\; \int \; -->}_{C_1}^{}\stackrel{<!--\; \rightharpoonup \; -->}{F}\cdot d\stackrel{<!--\; \rightharpoonup \; -->}{r}$

${\int <!--\; \int \; -->}_{C_1}^{}\stackrel{<!--\; \rightharpoonup \; -->}{F}\cdot d\stackrel{<!--\; \rightharpoonup \; -->}{r}$
$\le {\int <!--\; \int \; -->}_{C_3}^{}\stackrel{}{F}\cdot d\stackrel{<!--\; \rightharpoonup \; -->}{r}$
$\le {\int <!--\; \int \; -->}_{C_2}^{}\stackrel{<!--\; \rightharpoonup \; -->}{F}\cdot d\stackrel{<!--\; \rightharpoonup \; -->}{r}$
$\le {\int <!--\; \int \; -->}_{C_4}^{}\stackrel{<!--\; \rightharpoonup \; -->}{F}\cdot d\stackrel{<!--\; \rightharpoonup \; -->}{r}$

${\int <!--\; \int \; -->}_{C_1}^{}\stackrel{<!--\; \rightharpoonup \; -->}{F}\cdot d\stackrel{<!--\; \rightharpoonup \; -->}{r}$
$={\int <!--\; \int \; -->}_{C_2}^{}\stackrel{}{F}\cdot d\stackrel{<!--\; \rightharpoonup \; -->}{r}$
$={\int <!--\; \int \; -->}_{C_3}^{}\stackrel{<!--\; \rightharpoonup \; -->}{F}\cdot d\stackrel{<!--\; \rightharpoonup \; -->}{r}$
$={\int <!--\; \int \; -->}_{C_4}^{}\stackrel{<!--\; \rightharpoonup \; -->}{F}\cdot d\stackrel{<!--\; \rightharpoonup \; -->}{r}$

It is possible because the line integrals along all four paths are equal when $t=0.$

It is possible because the line integrals are equal along all four paths from $\left(0,0\right)$ to $\left(1,1\right)$.

No because the line integrals are not all equal when $t=0.5.$

There is not enough information because there could be other paths where the line integrals have equal values when $t=0.5.$

Yes because ${\int <!--\; \int \; -->}_{C_{}}^{}\stackrel{<!--\; \rightharpoonup \; -->}{F}\cdot d\stackrel{<!--\; \rightharpoonup \; -->}{r}=0$ for all circles in $D$

Yes because ${\int <!--\; \int \; -->}_{C_{}}^{}\stackrel{<!--\; \rightharpoonup \; -->}{F}\cdot d\stackrel{<!--\; \rightharpoonup \; -->}{r}=0$ for all closed paths in $D$

No because when the curve contains $\left(0,0\right)$ in its interior, ${\int <!--\; \int \; -->}_{C_{}}^{}\stackrel{<!--\; \rightharpoonup \; -->}{F}\cdot d\stackrel{<!--\; \rightharpoonup \; -->}{r}\ne 0$

There is not enough information. Even though ${\int <!--\; \int \; -->}_{C_{}}^{}\stackrel{<!--\; \rightharpoonup \; -->}{F}\cdot d\stackrel{<!--\; \rightharpoonup \; -->}{r}=0$ for all the closed paths shown, there could be other closed paths in D for which ${\int <!--\; \int \; -->}_{C_{}}^{}\stackrel{<!--\; \rightharpoonup \; -->}{F}\cdot d\stackrel{<!--\; \rightharpoonup \; -->}{r}\ne 0$