<s>[INST] Impossible integral? 
$$\int_{0}^{x^2-1} f(t) dt= x^6+x^4+3x^2$$

I saw this problem in a calculus exam. $f$ is assumed to be continuous. Using the Fundamental Theorem of Calculus, I calculated $f(t)= 3t^2+8t+8 $. But when I integrate $f$ it gives me $\int_{0}^{x^2-1} f(t) dt= x^6+x^4+3x^2-5$:
$$\frac {d}{dx}\int_{0}^{x^2-1} f(t) dt= \frac {d}{dx}[x^6+x^4+3x^2]$$
$$f(x^2-1)2x = 6x^5+4x^3+6x$$
$$f(x^2-1) = 3x^4+2x^2+3$$
$$f(x^2-1) = 3(x^2-1)^2+8(x^2-1)+8$$
$$f(t) = 3t^2+8t+8$$
$$\int_{0}^{x^2-1} 3t^2+8t+8 dt= x^6+x^4+3x^2-5$$
This means that there is no continuous function that satisfies the equation of the problem?
 [/INST] You are correct that this does mean there are no continuous functions satisfying the equation. (There's a quicker way to see this: set $x=1$, then $LHS=0$, but $RHS=5$.)
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