Applying FTC to improper integrals When reviewing improper integrals, I came across the following explanation;
To find the integral
$$\int_{2}^{5}\frac{1}{\sqrt{x-2}} \,dx,$$
we have to proceed as
$$\int_{2}^{5}\frac{1}{\sqrt{x-2}} \,dx=\lim_{t\rightarrow2^{+}}\int_{t}^{5}\frac{1}{\sqrt{x-2}} \,dx=\lim_{t\rightarrow2^{+}}2\sqrt{x-2}\,\Big|_t^5
=2\sqrt{3}$$
since the function $f(x)=1/\sqrt{x-2}$ has a singularity at $x=2$.
But in this case, an antiderivative of $f$, $F(x)=2\sqrt{x-2}\,$, is continuous on $[2, 5]$, so can't we just use the FTC and say as follows?
$$\int_{2}^{5}\frac{1}{\sqrt{x-2}} \,dx=2\sqrt{x-2}\,\Big|_2^5=2\sqrt{3}$$
I am studying for the GRE subject test, and I saw a couple of other examples that tell you to use the limit for the limit of integration.
Another example is
$$\int_{0}^{\infty}\frac{e^{-\sqrt{x}}}{\sqrt{x}} \,dx.$$
Here, the explanation is that since $g(x)=e^{-\sqrt{x}}/\sqrt{x}$ has a discontinuity at $x=0$, we have to treat the lower limit as $$\lim_{a\rightarrow0^{+}}\int_{a}^{\infty}\frac{e^{-\sqrt{x}}}{\sqrt{x}} \,dx.$$
But an antiderivatie of $g$ is continuous at $x=0$, so I don't know why this is necessary.
 A: The FTC guarantees the existence of an antiderivative when the integrand is everywhere continuous. If the integrand has a singularity anywhere on the domain $[a,b]$ then it may or may not have an antiderivative continuously definable on $[a,b]$. If such antiderivative does exist, say $F$ of $f$, then (e.g.) $\lim_{c\to0^+}\int_c^1f=\lim_{c\to0^+}F(1)-F(c)=F(1)-F(0)$ since $F$ is continuous. All is well: using antiderivative is a valid way to compute these integrals for continuous integrands.
The trouble lies in how an integral is defined. In Riemann integration, $\int_a^bf$ only makes sense if $f$ is every defined and finite on $[a,b]$ (and integrable…). So, should there be any singularity, we make a new definition by taking limits around the singularities. Though antiderivative are useful for computation, we can’t use antiderivatives in the definition since they may not necessarily exist: we may want to integrate a non-continuous function, e.g.
Consider $f(x)=1/x$. On $(0,1]$ it has an antiderivative $\ln x$. But no value of $\ln(0)$ will make $\ln:[0,1]\to\Bbb R$ continuous! And, as one might expect from this, $\int_0^1f$ does not converge as an improper integral.
A: The anti-derivatives themselves may be defined at $2$ and $0$ respectively, but the definite integral concerns itself with the continuity of the integrand, and this carries over to the anti-derivative in this case. You can see that the anti-derivatives are not defined to the left of $2$ and $0$ respectively, and so we do indeed need to approach these with caution as well. However, upon computation, it may be evident that you can simply plug in the upper and lower limits, this will not be the case in general, as the anti-derivative need not be defined at the upper or lower limits.
I think the go-to example might be $\int_0^1 \frac{1}{x}dx.$ You must approach $0$ from the right. You see that $\ln|x|$ is not defined at $x=0,$ but it is defined for every $x>0.$ So without knowing the anti-derivative ahead of time we must be wary.
