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  1. Given that the continuous function $f: \Bbb R \longrightarrow \Bbb R$ satisfies $$\int_0^\pi f(x) ~dx = \pi,$$ Find the exact value of $$\int_0^{\pi^{1/6}} x^5 f(x^6) ~dx.$$

  2. 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.

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There are many many many functions $f$ such that $$\int_0^{\pi}f(x)\,dx=\pi$$ It is true that the constant function $f(x)=1$ is one of them, and that shows that if the question has an answer then you can get the answer by using that constant function. It is probably meant to be an exercise in integration by substitution, letting $u=x^6$ or something like that.

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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 $$f(x) = 2- \dfrac{2x}{\pi}$$ enter image description here

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

enter image description here 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 $$g'(t) = f(b(t)) b'(t) - f(a(t)) a'(t)$$

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