Derivatives vs Integration 
  
*
  
*Given that the continuous function $f: \Bbb R \longrightarrow \Bbb R$ satisfies
  $$\int_0^\pi f(x) ~dx = \pi,$$
  Find the exact value of
  $$\int_0^{\pi^{1/6}} x^5 f(x^6) ~dx.$$
  
*Let
  $$g(t) = \int_t^{2t} \frac{x^2 + 1}{x + 1} ~dx.$$
  Find $g'(t)$.

For the first question: The way I understand this is that the area under $f(x)$ from $0$ to $\pi$ is $\pi$. Doesn't this mean that the function can be $f(x)=1$? Are there other functions that satisfy this definition? The second line in part one also confuses me, specifically the $x^6$ part!
For the second question: Does this have to do something with the Second Fundamental Theory of Calculus? I see that there are two variables, $x$ and $t$, that are involved in this equation. 

For the first question,
There are infinitely many functions other than $1$ that satisfy 
$$\int_0^{\pi} f(x) dx = \pi$$
For instance, couple of other examples are $$f(x) = 2- \dfrac{2x}{\pi}$$

and $$f(x) = \dfrac{2x}{\pi}$$

To evaluate $$\int_0^{\pi^{1/6}} x^5 f(x^6) dx$$ make the substitution $t = x^6$ and see what happens...
For the second question, Yes make use of the fundamental theorem of calculus i.e. if
$$g(t) = \int_{a(t)}^{b(t)} f(x) dx$$ then
$$g'(t) = f(b(t)) b'(t) - f(a(t)) a'(t)$$
