Question

Question about "Expanding the integral."

Answer 1

http://www.wolframalpha.com/input/?i=integral+1%2F%28sqrt%28x%5E4%29%29

Answer 2

so I'm curious how we get things like 1/(sqrt(x^4) = sqrt(x^4)/x^4

Answer 3

or that sqrt(x^4)/x^4 = 1/x^2

Answer 4

Is this one approach to many for doing a question like this? Sometimes Wolfram does some "interesting" things...

Answer 5

i dont know about the integral part, but sqrt(x^4)/x^4 = 1/x^2 because the square root of x^4 is x^2. and then the bottom can become x^2. so its 1/x^2

Answer 6

Ty Jessica.

Answer 7

are you familiar with the magic \[\frac{1}{\sqrt x} \implies \frac{1}{\sqrt x} \times \frac{\sqrt x}{\sqrt x} \implies \frac{\sqrt x}{x}\]

Answer 8

it follows the same principle

Answer 9

no.

Answer 10

you're not familiar with rationalization?

Answer 11

teach me about everything.

Answer 12

i find that very surprising...

Answer 13

hmm now that I look at that I think I've seen that before... So we do this with sqrt's usually?

Answer 14

rationalization is done when you have a radical in the denominator..your goal is to "neutralize" the denominator

and depends on the situation really

Answer 15

hmm cool trick :)

Answer 16

I have been denied this cool tricks my entire childhood.

Answer 17

for example you have \[\int \frac{x-1}{\sqrt x}dx\]
there's no need for rationalization...just split up the integral
\[\int \frac{x}{\sqrt x} - \int \frac{1}{\sqrt x}\]
you can use rationalization but you know this is just \[\int x^{1 - 1/2} - \int x^{-1/2}\]

Answer 18

\[x^{1 - 1/2} = x^{1/2}\]
now notice if i use rationalization
\[\frac{x}{\sqrt x} \implies \frac{x}{\sqrt x} \times \frac{\sqrt x}{\sqrt x} \implies \frac{x\sqrt x}{x} \implies \sqrt x \implies x^{1/2}\]

Answer 19

both will result the same answers

Answer 20

so if you had a x^1/2 you could randomly do x/(x)^(1/2)

Answer 21

you're soo cool iggy.

Answer 22

TEACH ME MOAR.

Answer 23

@lgbasallote run away after medal eh?

Answer 24

lol i ran away much earlier :p

Answer 25

and why would you make x^1/2 into a more complicated state?

Answer 26

i guess if you have some outside of the box plan...it's possible

Answer 27

I'm getting this from another calculator

Answer 28

I'm sure there's something to do with "restricted values of x" in wolfram

Answer 29

wolfram

Answer 30

:)