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Let $g(x)$ be a continuous function for any $x \in R$ that satisfies:

$$2x^5 + 64 = \int_c^x g(t) \,dt$$

Find the value of $c$ or prove it does not exist.


I suppose the starting point is the following:

$$2(x^5 + 2^5) = \int_c^x g(t) \,dt$$ $$x^5 = \frac{\int_c^x g(t) \,dt}{2} - 2^5$$ $$x^5 = \frac{G(x) - G(c)}{2} - 2^5$$

But what should follow afterwards?

Thanks in advance.

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Well put $x=c$ in your equation, which will give $\int_c^{c} g(t)dt = 0$. You will simply get $-2c^5 = 64 \iff c = -2$.

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  • $\begingroup$ Hahah I was mistaken anyway. The term $2x^5$ cancels out on both sides of the equation, leaving your answer: $c = -2$. You are correct :) $\endgroup$ – Mr Pie Apr 7 '18 at 10:00
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    $\begingroup$ @user yes i understand, nevermind! $\endgroup$ – King Tut Apr 7 '18 at 10:20
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By the Fundamental Theorem of Calculus we have \begin{align*} (2x^{5}+64)'=g(x), \end{align*} then $g(x)=10x^{4}$. Substituting back to the expression we get \begin{align*} 2x^{5}+64=\int_{c}^{x}10t^{4}dt=2t^{5}\bigg|_{t=c}^{t=x}=2x^{5}-2c^{5}, \end{align*} so $c^{5}=-32$ and hence $c=-2$.

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