The problem given is:

$\int_3^5 f(x) \, dx = \frac{35}{11}$

You are asked to find:

$\int_5^3 f(t) \, dt$

Notice that the second integral is the same as the first one but with reversed limits of integration. The property of definite integrals allows us to express the reversed integral as:

$\int_5^3 f(t) \, dt = - \int_3^5 f(t) \, dt$

Thus, we can substitute the known value of the first integral:

$\int_5^3 f(t) \, dt = - \frac{35}{11}$

So, the answer is:

$\boxed{-\frac{35}{11}}$

Would you like more details on this, or do you have any questions?

Here are five related questions for further practice:

1. What is the geometric interpretation of a definite integral with reversed limits?
2. If $\int_3^5 f(x)dx = \frac{35}{11}$, what is $\int_3^5 2f(x)dx$?
3. How would you solve $\int_3^5 f(x)dx = k$ for $\int_5^3 f(x)dx$ in general terms?
4. How do you calculate definite integrals for more complex functions, like $\int_3^5 (x^2 + 1)dx$?
5. What does the Fundamental Theorem of Calculus say about definite integrals?

Tip: Always remember, switching the limits of a definite integral changes its sign!