Question

What is \(u\)-substitution?

Answer 1

\[\int f(x)dx= \int udu?\]
tha one?

Answer 2

maybe not

Answer 3

Yeah, but I have seen \(u\)-substitution in normal equations as well...

Answer 4

Let \(u = \text{blah}\)

Answer 5

in normal equations that woud be algebraic substitution...

Answer 6

u-sub is just the term used in integration

Answer 7

OK, can I have an example?

Answer 8

\[\int x^5\text dx\]
let\[x^5=u\]\[x={u}^{1/5}\]\[\text dx=\frac{-4}{5}u^{1/5}\text du\]
\[\longrightarrow\int u\cdot\frac{1}{5}{u^{1/5}}\text du\]\[=\frac{1}{5}\int {u^{1/5}}\text du\]\[=\frac{1}{5}\frac{5u^{6/5}}{6}+c\]\[=\frac{1}{6}u^{6/5}+c\]\[=\frac{x^6}{x}+c\]

Answer 9

I don't know Calculus, but still... :\

Answer 10

well why did i spend all this time explaining ing u substitution? \[\int ax^n\text dx\]
let
\[ax^n=u\]\[x=\left(\frac{u}{a}\right)^{1/n}=\frac{u^{1/n}}{a^{1/n}}\]
\[\text dx=\frac{u^{1/n-1}}{na^{1/n}}\text du\]

\[\int ax^n\text dx\longrightarrow\frac{1}{na^{1/n}}\int u\cdot{u^{1/n-1}}\text du\]
\[=\frac{1}{na^{1/n}}\int {u^{1/n}}\text du\]\[=\frac{1}{na^{1/n}}\frac{u^{1/n+1}}{1/n+1}+c\]\[={}\frac{a^{1+n}x^{1+n}}{a^{1/n}+na^{1/n}}+c\]\[=\frac{ax^{1+n}}{1+n}+c\]

Answer 11

Is Calculus too hard?

Answer 12

I just wanted to get the idea. Sorry for the inconvenience. Wait, let me reward you.

Answer 13

What is an example of a normal equation that has \(u\)-substitution

Answer 14

\[ x^4 + 10x^2 + 25 = 0\]Let \( u = x^2\)\[ u^2 + 10u + 25 = 0\]\[ u=-5\]

Answer 15

But as I learnt by LG's reply, this is algebraic substitution.

Answer 16

where is the confusion
\[x=\pm\sqrt{-5}=\pm\sqrt 5i\]

Answer 17

Yeah... lol... I do know that.

Answer 18

substitution is substitution

Answer 19

Heh, thank you so much!