How can I find Limit of integral
How can I find:
$$\lim_{x\to0} \int_{0}^{x} e^{-t}\mathrm{d}t$$
I think to use fundamental method of calculus?
$F(x)= \int_{0}^{x} e^{-t}\mathrm{d}t$
Then
$F'(x)=-e^{-t}$
But I don't get any result!
Then I think to integrate then take a limit
$$ \int_{0}^{x} e^{-t}\mathrm{d}t= 1-e^{-x}$$
$$\lim_{x\to0} \int_{0}^{x} e^{-t}\mathrm{d}t=\lim_{x\to0} (1-e^{-x})=??$$
But also I don't know how can I find this limit.
$\endgroup$ 43 Answers
$\begingroup$You can prove that an even stronger result is true. Specifically that if $f$ is an integrable function, then
$$\lim_{x\rightarrow a} \int_{a}^xf(t)dt=0$$
Indeed, from the Second Fundamental Theorem of Calculus it holds that if $G$ is an antiderivative of $f$, then $$\int_{a}^xf(t)dt=G(x)-G(a)$$
But $G$ is a continuous function, so you obtain $$\lim_{x\rightarrow a}\Big[G(x)-G(a)\Big]=G(a)-G(a)=0$$
$\endgroup$ 1 $\begingroup$You actually don't need to do anything. Functions of the form $$h(x) = \int_a^x f(t) \ dt$$
where $f$ is integrable are continuous. The result follows trivially.
$\endgroup$ 0 $\begingroup$First evaluate the integral. This is done by subtracting the upper bound from the lower bound in the indefinite integral. I.E. Second Fundamental Theorem.
This yields:
$$-1 + e^{-x}$$
Then we wish to find the limit as it goes to zero. Here's a nice fact. Even when an integral has infinite area it is always continuous where it is defined. In other words, the value approaching zero is the value at zero. This is why divergent and improper integrals "split" the integral as splitting can be rewritten as an equation to remove jump discontinuity. I.E. The main problem with discontinuous/asymptotic function integrals. In fact, such an equation uses the limits of integration... As actual limits! The only reason this changes is when the indefinite integral that was calculated is discontinuous (sometimes they are easier to calculate due to the splitting being messy).
Anyway enough rambling...
The answer is:
$$-1 + e^0 = -1 + 1 = 0$$
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