Improper Integral $\int_0^2 \frac{1}{\sqrt{x}} \ \text{d}x$

Question

I have to find the value of: $$\int_0^2 \dfrac{1}{\sqrt{x}} \ \text{d}x$$ Here is my work so far: $$\int_0^2 \dfrac{1}{\sqrt{x}} \ \text{d}x$$ $$=\int_0^2 x^{-1/2} \ \text{d}x$$ $$=\left.2x^{1/2}\right|_0^2$$ At this point I was about to apply the Fundamental Theorem, then I realized that the function is discontinuous at $x=0$, so the integral does not exist. However, the answer is $2\sqrt{2}$, which is what I would have gotten if the Fundamental Theorem of Calculus could have been applied. First, how would I get $2\sqrt{2}$, and second, is it simply a coincidence that the answer is $2\sqrt{2}$?

Answer 1

The function $1/\sqrt{x}$ is not defined at $0$, so it's an abuse of notation to write $$\int_0^2 \dfrac{1}{\sqrt{x}} \ dx$$

You're really being asked to compute the improper integral $$ \lim_{t \to 0^+} \int_t^2 \frac{1}{\sqrt{x}} dx$$