Find $c$ in $2x^5 + 64 = \int_c^x g(t) \,dx$

Question

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.

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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.

Answer 1

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

Answer 2

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 $c=-2$.