Question

Midterm tmrw, please help!? Integral substitution?

Answer 1

which integral?
or do you want to learn about u-substitution in integration in general ?

Answer 2

\[\int\limits_{?}^{?}10x(5x^2-3)\]

Answer 3

what is the derivative of 5x^2-3 ?

Answer 4

Ignore the question marks I didn't know how to get rid of them. Now I just have a question about this problem!

Answer 5

I understand how I would use substitution to integrate the (\[(5x^2-3)\] But what happens to the 10x??

Answer 6

that it's being multiplied by

Answer 7

\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\color{red}{10x}\color{blue}{\left(5x^2-3\right)}~\color{red}{dx}}\)

you set: \(\large\color{slate}{\displaystyle u=\color{blue}{5x^2-3}}\)
and \(\large\color{slate}{\displaystyle du=\color{red}{10x~dx}}\)

this gets you,
\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\color{blue}{\left(u\right)}~\color{red}{du}}\)

Answer 8

the dx times the derivative of u (in this case 10x) is all replaced by du
and u-substitution is just as it is, whatever we substitute for u, that we replace.

Answer 9

@Anon101

Answer 10

:D

Answer 11

pita, are you fine with what I posted ?

Answer 12

How did you find du again?

Answer 13

du = dx times [the derivative of what you are substitution for u]

Answer 14

So you don't integrate the 10x at all that you highlighted in red in the initial equation?

Answer 15

well, 10x goes away as we substitute

Answer 16

Again, for review

\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\color{green}{10x}\color{blue}{\left(5x^2-3\right)}~\color{green}{dx}}\)

you set: \(\large\color{slate}{\displaystyle u=\color{blue}{5x^2-3}}\)
and \(\large\color{slate}{\displaystyle du=\color{green}{10x~dx}}\)

this gets you,
\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\color{blue}{\left(u\right)}~\color{green}{du}}\)

Answer 17

Shoot I'm really sorry abou tthis. but the equation was \[10x \int\limits_{?}^{?}(5x^2-3)^6\]

Answer 18

you will apply the power rule to the integral of u du, and then substitute back 5x^2-3 instead of u
(and don't forget the +C )

Answer 19

I missed the power of 6^

Answer 20

it is pretty much same

Answer 21

I understand that it would be (5x^2-3)^7 / 7 +C

Answer 22

Just not why the 10x before it disappeared

Answer 23

\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\color{red}{10x}\left(\color{blue}{5x^2-3}\right)^6~\color{red}{dx}}\)

you set: \(\large\color{slate}{\displaystyle u=\color{blue}{5x^2-3}}\)
and \(\large\color{slate}{\displaystyle du=\color{red}{10x~dx}}\)

this gets you,
\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\left(\color{blue}{u}\right)^6~\color{red}{du}}\)

Answer 24

See how the u sub works?

Answer 25

Basically that:

\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\color{red}{g'(x)}~~F\left(~\color{blue}{g(x)~}\right)~\color{red}{dx}}\)

you set: \(\large\color{slate}{\displaystyle u=\color{blue}{g(x)}}\)
and \(\large\color{slate}{\displaystyle du=\color{red}{f'(x) ~dx}}\)

this gets you,
\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}F\left(\color{blue}{u}\right)~\color{red}{du}}\)

Answer 26

so no matter what power you had there, be it 6 or -9 or any k
then we would be integrating u^k du

Answer 27

Sorry, but where does the g'(x) go when you are integrating?

Answer 28

it goes away:

u is replacing some function g(x) (instead of which you are substituting u)
du is replacing the ( `g'(x) times dx` )

Answer 29

if u=5x^2-3
then what is du/dx equal to @pita6602 ?
Have you done derivatives before?

Answer 30

Yeah no I got it. Thank you Solomon, that helped a lot!

Answer 31

So from now on I can just ignore g'(x). But i get it.

Answer 32

\[u=5x^2-3 \\ \frac{du}{dx}=\frac{d(5x^2-3)}{dx}=10x \\ \\ \frac{du}{dx}=10 x \\ du=10x dx\]
well 10x doesn't just go away
it is just a factor of du

Answer 33

well, once everything in that form of: \(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\color{red}{g'(x)~}~F\left(\color{blue}{~g(x)~}\right)~\color{red}{dx}}\)

then,

you set: \(\large\color{slate}{\displaystyle u=\color{blue}{g(x)}}\)
and \(\large\color{slate}{\displaystyle du=\color{red}{g'(x) ~dx}}\)

and then this will get you
\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}~F\left(\color{blue}{u}\right)~\color{red}{du}}\)

Answer 34

obviously not everything in that form, but in this case it is so.

Answer 35

\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\color{red}{10x}\left(\color{blue}{5x^2-3}\right)^6~\color{red}{dx}}\)

you set: \(\large\color{slate}{\displaystyle u=\color{blue}{5x^2-3}}\)
and \(\large\color{slate}{\displaystyle du=\color{red}{10x~dx}}\)

this gets you,
\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\left(\color{blue}{u}\right)^6~\color{red}{du}}\)

Answer 36

then finish integrating it
\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\left(\color{blue}{u}\right)^6~\color{red}{du}=\frac{\color{blue}{u}^{6+1}}{6+1}+C=\frac{u^7}{7}+C}\)

Answer 37

and then substitute 5x^2-3 back for u