Given that $$\int_{c}^xf(t)\,\mathrm{d}t=x^3 + x^5,$$ where $c$ is constant, find the value of $c$.
I started by getting finding $D_x(\int_{c}^xf(t)\,\mathrm{d}t)$:
$$D_x\left(\int_{c}^xf(t)\,\mathrm{d}t\right)=D_x(x^3 + x^5) = 3x^2 + 5x^4.$$
Thus we can say that $f(x)$ $=$ $3x^2$ $+$ $5x^4$.
I am stuck after this step. Kindly help me. Thanks in advance.
Realize that if $c \neq 0$, there would be some constant part, like so:
$$\int_{c}^x f(t)dt = x^3 + x^5 - c^3 - c^5$$
But since there is no constant part, you can deduce that $c = 0$. You can show a bit clearer, applying the Fundamental Theorem of Calculus as you did:
$$ \require{cancel} \dfrac{d}{dx}\int_c^x f(t)dt = \dfrac{d}{dx}\left(x^3+x^5\right) \Rightarrow f(t) = 3t^2 + 5t^4$$ $$\int_c^x \left(3t^2 + 5x^4\right) dt = \left(x^3 + 5x^4\right) - \left(c^3 + c^5\right)$$ $$x ^3 + x^5 = \left(x^3 + x^5\right) - \left(c^3 + c^5\right)$$ $$0 = - \left(c^3 + c^5\right)$$ $$0 = -c^3\cancel{\left(1+c^2\right)}$$ $$c = 0$$