0
$\begingroup$

I have recently started an integration unit, and im just now getting familiar eith the concept of integrals. However, I cant figure out why $$\int\frac{1}{\sqrt{x}}dx=2\sqrt{x}+C$$ Wolfram Alpha doesnt show any steps for it, and I cant seem to work it out on paper. Any help would be appreciated!

$\endgroup$
  • 2
    $\begingroup$ What is the derivative of $2\sqrt x+C$? $\endgroup$ – ℋolo Aug 28 '18 at 11:16
  • 1
    $\begingroup$ $\frac{1}{\sqrt{x}}=\frac{1}{x^{1/2}}=x^{-1/2}$ $\endgroup$ – Botond Aug 28 '18 at 11:16
  • $\begingroup$ @Holo got it, but how am i supposed to know that, unless i see the answer? or is this just one of those properties that i have to memorize? $\endgroup$ – Pablo Aug 28 '18 at 11:17
  • 2
    $\begingroup$ @Pablo you can know the derivatives of $x^r$ for $r\in \Bbb R$, and then deduce that since the integrand is $x^{-1/2}$, then the integral must be $A\cdot \sqrt x + C$ for some constants $A, C$. Finding $A$ after that isn't too hard. $\endgroup$ – Arthur Aug 28 '18 at 11:19
  • 1
    $\begingroup$ @Pablo not at all, Dr. Sonnhard Graubner's answer shows how to get there, this is just a way to confirm things. Also note that you forgot $dx$ in the integral $\endgroup$ – ℋolo Aug 28 '18 at 11:19
3
$\begingroup$

The power rule says that the derivative of $x^r$ with respect to $x$ is

$rx^{r-1}$

Knowing that $$\frac{1}{\sqrt{x}}=x^{-\frac{1}{2}}$$

The derivative of $\sqrt{x}+C$ is, by the power rule

$$\frac{1}{2}x^{-\frac{1}{2}}$$

So the derivative of $2\sqrt{x}+C$ is $

$$x^{-\frac{1}{2}}=\frac{1}{\sqrt{x}}$$

and so the integral is as given.

$\endgroup$
5
$\begingroup$

You can use that $$\int x^ndx=\frac{x^{n+1}}{n+1}+C$$ if $$n\ne -1$$ and write your integral as

$$\int x^{-1/2}dx$$

$\endgroup$
  • $\begingroup$ alright got it. thank you very much! $\endgroup$ – Pablo Aug 28 '18 at 11:19
2
$\begingroup$

One of the first integration trick is to figure out what function can give you the desired integrand. In many cases both functions have similarities, so what you can do is to differentiate the integrand and try to find an "invertible" pattern.

In the case on hand,

$$\left(\frac1{\sqrt x}\right)'=(x^{-1/2})'=-\frac12x^{-3/2}$$ and that reminds you that the derivative of a power is also a power, namely with the initial exponent minus one.

Now try $$(x^{1/2})'=\frac12x^{-1/2}=\frac1{2\sqrt x}$$

and you are very close.

$\endgroup$
0
$\begingroup$

For fun, Integration by substitution:

$x= \phi (t) =t^2$ , then

with $f(x)= \frac{1}{√x}$,

$\int f(x)dx = \int f(\phi(t)) \phi'(t)dt$, or

$\int \frac{dx}{√x} = \int (\frac{1}{t})(2t) dt=$

$2 \int dt =2t = 2√x.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.