Assuming that $F(x)$ is continuous ,what is the derivation of integral

Question

Assuming that $F(x)$ is continuous ,what is the derivation of $$\int_{1}^x (x^2+f(t)) \,dt \;?$$ we know that answer is not $x^2+f(x)$.

Answer 1

Let $F(x)$ be an antiderivative of $f(x)$.

\begin{align} \frac{d}{dx} \int_1^x (x^2+f(t))\,dt & = \frac{d}{dx} \int_1^x x^2\,dt+\frac{d}{dx} \int_1^x f(t)\,dt \\ & = \frac{d}{dx}\left(x^2\int_1^x \,dt\right)+\frac{d}{dx} \int_1^x f(t)\,dt\\ & = \frac{d}{dx}\left(x^2(x-1)\right)+\frac{d}{dx} \left(F(x)-F(1)\right)\\ & = 3x^2-2x+f(x) \end{align}

Answer 2

It is equal to $x^3+f(x)$ through the fact that $x$ is constant relative to integral variable $t$. use the linear property of integral and Newton theorem of derivative of an integral.

Answer 3

let $g(x) =\int_{1}^x (x^2+f(t)) \,dt$, you are looking for $g'(x)$.

Note that $g(x)=\int_{1}^x x^2 \,dt + \int_{1}^x f(t) \,dt = x^2(x-1) + \int_{1}^x f(t) \,dt$, therefore, $g'(x)=3x^2-2x+f(x)$