Let $f$ be a differentiable function on $[0,1]$.Is the indefinite integral of $f$ differentiable on $[0,1]$?
By the fundamental theorem of calculus i know that the integral will be differentiable in $(0,1)$.But what about differentiability at the points 0 and 1?
On
$f$ is continuous on $[0,1]$, hence $f$ has an antiderivative $F$ on $[0,1]$.
Differentiability of $F$ at the points $0$ and $1$ is defined by
$\lim_{x \to 0+0}\frac{F(x)-F(0)}{x-0}$ resp. $\lim_{x \to 1-0}\frac{F(x)-F(1)}{x-1}$.
It holds that $\lim_{x \to 0+0}\frac{F(x)-F(0)}{x-0}=f(0)$ and $\lim_{x \to 1-0}\frac{F(x)-F(1)}{x-1}=f(1)$.
If $G$ is an indefinite integral of $f$, then there is a constant $C$ such that $G=F+C$ on $[0,1]$. Hence $G$ is differentiable on $[0,1]$.
First of all, check out the fact that continuity of $f$ is enough for differentiability of a primitive. Of course, differentiability is even stronger so it works.
If by 'indefinite integral' of a function $f$ you mean one of its primitives (some refer instead to the whole family of these primitives by that name) then the answer is yes. Let's call it $F(x)=\int_a^x f(t)$ for some $a\in [0,1]$ and let's see that $F$ is also differentiable in the extreme points of $[0,1]$ in the same sense that $f$ is.
Let's say that '$f$ is differentiable in $[0,1]$' means that there exists a differentiable function $\tilde f$ over an open interval $\mathcal I\supset[0,1]$ whose restriction to $[0,1]$ coincides with $f$. Then by applying the Fundamental Theorem of Calculus to $\tilde f$ we get that $$\tilde F(x)=\int_a^x \tilde f(t) dt$$ is well defined and differentiable for every $x\in\mathcal I$, and in particular $$F(x)=\int_a^xf(t)dt=\int_a^x \tilde f(t) dt=\tilde F(x)$$ for $x \in [0,1]$. This shows that $F$ is differentiable in $[0,1]$ in the postulated sense.
Equivalently you can define existence of derivative at $0$ or $1$ as the existence of the respective lateral derivatives and you would arrive to the same conclusion.