calculate a generalized integral

Question

I want to calculate a generalized integral:

$$\int^1_0\frac{dx}{\sqrt{1-x}}$$

I have a theorem :

if $f(x)$ is continuous over $[a,b[$ then:

$$\int^b_af(x).dx = \lim_{c\to b⁻}\int^c_af(x).dx$$

if $f(x)$ is continuous over $]a,b]$ then:

$$\int^b_af(x).dx = \lim_{c\to a⁺}\int^b_cf(x).dx$$

if $f(x)$ is continuous over $[a,b]$ except at the point $c$ in the $]a,b[$:

$$\int^b_af(x).dx = \lim_{c\to b⁻}\int^c_af(x).dx + \lim_{c\to a⁺}\int^b_cf(x).dx$$

Then I studied the area of ​​its definition, which is: $D_f = ]1, -\infty[$

So I don't know where is the case of my integral.

could you help me please with that ?

Answer 1

Your integrand is continuous on the domain $[0,1)$. Hence, you need to evaluate your integral as follows: $$I=\int_0^1 \dfrac{dx}{\sqrt{1-x}} = \lim_{t \to 1^-}\int_0^t \dfrac{dx}{\sqrt{1-x}}$$ Now for $t<1$, setting $\sqrt{1-x} = u$, we have $1-x = u^2 \implies -dx = 2u du$. Hence, $$\int_0^t \dfrac{dx}{\sqrt{1-x}} = \int_1^{\sqrt{1-t}} \dfrac{-2udu}{u} = 2\int_{\sqrt{1-t}}^1 du = 2 \left(1-\sqrt{1-t}\right)$$ Now $$I = \lim_{t \to 1^-}2 \left(1-\sqrt{1-t}\right) = 2$$

Answer 2

There is a problem near $x=1$. So we find $$\int_0^{1-\epsilon} \frac{dx}{\sqrt{1-x}}\tag{$1$}$$ and then calculate the limit of this as $\epsilon$ approaches $0$ through positive values. Note that an antiderivative of $\frac{1}{\sqrt{1-x}}$ is $-2\sqrt{1-x}$. It follows that the integral $(1)$ is equal to $$2(1-\sqrt{\epsilon}).$$ As $\epsilon$ approaches $0$ through positive values, this approaches $2$. More formally, $$\lim_{\epsilon\to 0^+}\int_0^{1-\epsilon} \frac{dx}{\sqrt{1-x}}=\lim_{\epsilon\to 0^+}2(1-\sqrt{\epsilon})=2,$$ so our improper integral converges, and has value $2$.