Question

I need help understanding u substitution :)

Answer 1

Well do you know you need to also write dx in terms of u and du ?

Answer 2

Yeah. Do I have to come up with du then?

Answer 3

We did this towards the end of class and I couldn't fully understand haha xD

Answer 4

The U is just the inside material. It is just helping you more clearly visualize the inside and out side areas for more complex problems.

Answer 5

du/dx is just the derivative of u with respect to x

Answer 6

\[u=7+x^4 \\ \frac{du}{dx}=(7+x^4)' \\ du=(7+x^4)' dx\]

Answer 7

but that is just everything inside of the parenthesis?

Answer 8

yep aka the thing they called u

Answer 9

you do know (u)' means to find the derivative of u?

Answer 10

right yeah, I know that.
But then what about the x^3 and ^4 around all that o.o

Answer 11

we will get there
have you differentiated 7+x^4 yet?

Answer 12

:0

Answer 13

darn integrals xD

Answer 14

du = 4x^3

Answer 15

\[u=7+x^4 \\ \frac{du}{dx}=(7+x^4) ' \\ \frac{du}{dx}=0+4x^3 \\ \frac{du}{dx}=4x^3 \\ du=4x^3 dx \\ \\ \text{ or dividing 4 on both sides } \\ \frac{1}{4} du=x^3 dx \\ \text{ so you have } \\ \int\limits \color{red}{x^3} (\color{blue}{7+x^4)}^4 \color{red}{dx} =\int\limits \color{blue}{u}^4 \color{red}{\frac{1}{4} du}\]

Answer 16

do you know how to integrate u^4 with respect to u?

Answer 17

I know nothing xD

Answer 18

Do you know power rule for integration?

Answer 19

\[\int\limits u^n du=\frac{u^{n+1}}{n+1}+C , n \neq -1 \]

Answer 20

you have n is 4 here

Answer 21

soo..it's the same as normal.

Answer 22

yes... I think I think I know what you ...
\[\int\limits x^4 dx=\frac{x^5}{5}+C \\ \text{ similarly } \\ \int\limits u^4 du=\frac{u^5}{5}+C\]
is that what you mean?

Answer 23

yeah :)

Answer 24

we don't always have to integrate with respect to x

Answer 25

Right, I see.
so with that 1/4, do I just pull it out to the left of the integral sign thing?

Answer 26

another example
\[\int\limits \star d \star =\frac{\star^2}{2}+C\]

Answer 27

yes it is a constant multiple

Answer 28

just bring it down

Answer 29

\[\int\limits c f(x) dx=c \int\limits f(x) dx\]

Answer 30

and don't forget at the end to replace u with 7+x^4

Answer 31

so in the end this should equal..
(u^5)/20
= [(7+x^4)^5]/20 ?

Answer 32

+C

Answer 33

I replace the u after I finish sloving, yes?
oh yea, thanks

Answer 34

you may check answer by differentiating that answer and see if you get integrand:
\[\frac{d}{dx} \frac{1}{20}(7+x^4)^5 \\ \frac{1}{20} \frac{d}{dx}(7+x^4)^5 \\ \frac{1}{20}
(7+x^4)' \cdot 5(7+x^4)^4 \\ \frac{1}{20}(4x^3) \cdot 5(7+x^4)^4 \\ \frac{20}{20}x^3 (7+x^4)^4 \\ 1 x^3(7+x^4) \\ x^3(7+x^4)\]

Answer 35

Awesome, thank you so much!!
now i'm off try another more difficult one xD

Answer 36

k you can write here if you want

Answer 37

\[\int\limits f'(x) \cdot (f(x))^n dx\]
usually in this form they go for u=f(x)
since du/dx=f'(x)
and you would have
\[\int\limits u^n du\]
there might be a constant multiple involved like with the one you just asked

Answer 38

oops, didn't see this until now!
the new one is
\[\int\limits_{}^{}\frac{ dt }{ (1-9t)^4 }\]

Answer 39

u = 1-9t

Answer 40

@myininaya

Answer 41

du = -9dx
then?

Answer 42

du=-9 dt

Answer 43

pr divide both sides by -9
-1/9 du = dt

Answer 44

and we end up with
(-1/9) int u^(-4)

Answer 45

yep
\[\frac{-1}{9} \int\limits u^{-4} du\]

Answer 46

= \[\frac{ 1 }{ 27(1-9t)^3 }+C\]

Answer 47

well done!

Answer 48

Thank you so much:)

Answer 49

np
goodnight

Answer 50

sleep happy!