Find a continuous function $f$ in accordance with FTC 
Find a continuous function $f$ and a constant $K$ such that $$\int_0^x f(t)dt = 2x^3\sin(x)+3e^{5x}+K$$

By the FTC I, we know that for a Riemann integrable function $f$ on $[a,b]$, where $a\leq x\leq b$ and $F(x) = \int_a^x f(t)\mathrm dt$, $F$ is differentiable at $x_0$ and $F'(x_0)=f(x_0)$. In other words, $$ \frac{d}{dx} \int_a^x f(t)\mathrm dt=f(x)$$ for all continuous points of $f$.
In this problem, we can let $F(x) = 2x^3\sin(x)+3e^{5x}+K$. Then, we have $F'(x) = 2x^3\cos(x)+6x^2\sin(x)+15e^{5x} =f(x)$.
However, I'm not sure if this is right. Is the answer this simple? Can I let the constant $K$ be any real number? Any hints would be greatly appreciated. Thanks in advance.

As you said, FTC tells you that $f$ is the derivative of the right side, which you can just calculate. With a completely indefinite integral (neither limit is specified), $K$ is arbitrary, but since the lower limit is fixed, $K$ is not arbitrary in your case. But if your original function is $F(x)$, then certainly $F(0)=0$, so you should have $2 \cdot 0 + 3 \cdot 1 + K = 0$.
