Solve integral weird upper bound approaching zero. I am trying to solve an integral of the form
$$p(0,t+\delta t)=\int_0^{\mu \delta t} f(x,t) dx$$
for $\delta t \rightarrow 0$ 
Intuitively, I would think that this integral has an upper bound which approaches the lower bound and therefore the result should be $f(0,t)$. However, this does not seem to be correct but I don't know how to approach the problem correctly. Has anybody a hint?
Update:
The result of this integral is supposed to be 
$$ \mu f(0,t) $$
Unfortunately, I do not see how this is obtained.
 A: It may be easier to see if we get rid of some notation. 
Replace $\delta t$ by the symbol $h$. Let $g(h)$ be the function 
$$ g(h) = p(0,t+h) = p(0,t + \delta t) $$
Let $k(x) = f(x,t)$. (Notice that throughout the question, $t$ is an arbitrary, but unchanging variable, so we can suppress it from the notation.) 
Then your question asks to find, as $h\to 0$, the quantity
$$ g(h) = \int_0^{\mu h} k(x) ~\mathrm{d}x \tag{1}$$
If $k(x)$ is bounded near $x = 0$ and integrable, then clearly 
$$ \lim_{h \to 0} g(h) = 0 $$
However, given the "expected answer", what you are after is more likely
$$ \lim_{h \to 0} \frac{g(h)}{h} $$
And given the more transparent form in equation (1), you see that by the fundamental theorem of calculus 
$$ \lim_{h \to 0} \frac{g(h)}{h} = g'(0) = \mu k(0) $$
as desired. 
The $\mu$ factor comes in as follows: let $K(s) = \int_0^s k(x) ~\mathrm{d}x$ be the antiderivative of $k$. Then we have 
$$ g(h) = K(\mu h) $$
So the $\mu$ factor appears from the chain rule when taking derivatives of both sides. 
