$\int_{0}^{1}\frac{x}{\sqrt{1-x}}dx$

Question

Solvable fixing $1-x=t$, I have a doubt about the integration's extremes. If $x=1\rightarrow t=1-x=0$, while if $x=0\rightarrow t=1-x=1$, so we have $-\int_{1}^{0}\frac{1-t}{\sqrt{t}}dt=\int_{0}^{1}\frac{1-t}{\sqrt{t}}dt$? Thanks in advance.

Answer 1

To interchange the limits on the integral, reverse the sign. $$ -\int_1^0\frac{1-t}{\sqrt{t}} dt = \int_0^1\frac{1-t}{\sqrt{t}} dt $$

Answer 2

Hint: Substitute $\sqrt{1-x}=t\implies x=1-t^2$ and $\displaystyle\frac{dx}{\sqrt{1-x}}=-2dt$

Answer 3

Yes, that’s correct. Sometimes, it is better not to change the limits and evaluate the anti-derivative and then at last put the original variable instead of substituted ones and then fill up the limits.

For example, your problem can be written like this: $$ -\int_{x=0}^{x=1} \frac{1-t}{\sqrt t} dt $$

$$-\int_{x=0}^{x=1} t^{-1/2} dt + \int_{x=0}^{x=1} t^{1/2} dt $$ $$-2\sqrt{t}\bigg|_{x=0}^{x=1} + \frac{2}{3}t^{3/2}\bigg|_{x=0}^{x=1}$$ $$-2 (1-x)^{1/2}\bigg|_{0}^{1} + \frac{2}{3} (1-x)^{3/2}\bigg|_{0}^{1} $$ $$ 2 -2/3= 4/3$$

Answer 4

The integral

$$ \int_0^1 \frac{x}{\sqrt{1-x}}\:dx $$

is an improper integral because the function

$$ f(x) = \frac{x}{\sqrt{1-x}} $$

is unbounded in $(1-\epsilon,1)$ for any $\epsilon>0$.

Therefore,

$$ \int_0^1 \frac{x}{\sqrt{1-x}}\:dx := \lim_{\epsilon \to 0} \int_0^{1-\epsilon} \frac{x}{\sqrt{1-x}}\:dx. $$

The substitution $t=1-x$ gives

$$ \begin{eqnarray*} \int_0^{1-\epsilon} \frac{x}{\sqrt{1-x}}\:dx & = & \int_{\epsilon}^1 \frac{1-t}{\sqrt{t}}\:dt \\ & = & \int_{\epsilon}^1 \left(\frac{1}{\sqrt{t}}-\sqrt{t}\right)\:dt \\ & = & \left[ 2\sqrt{t}- \frac{2}{3}t\sqrt{t} \right]_{\epsilon}^1 \\ & = & \frac{4}{3} - \left( 2\sqrt{\epsilon} - \frac{2}{3}\epsilon\sqrt{\epsilon} \right). \end{eqnarray*} $$

Taking limits as $\epsilon \to 0$ gives

$$ \int_0^1 \frac{x}{\sqrt{1-x}}\:dx = \frac{4}{3}. $$