Question

@DanJS

Answer 1

Interesting integral you got there: let me take a look

Answer 2

I will right it out for now,
\(\large\color{slate}{\displaystyle\int\limits_{-1}^{0}(2x^5+6x)^3(5x^4+3)dx}\)

Answer 3

here what you can do:

\(\large\color{slate}{\displaystyle\int\limits_{-1}^{0}2^3(x^5+3x)^3(5x^4+3)dx}\)

Answer 4

then you have an obvious u-substitution after taking the constant out

Answer 5

ok got it , whats next??

Answer 6

wait,

\(\large\color{slate}{\displaystyle\int\limits_{-1}^{0}2^3(x^5+3x)^3(5x^4+3)dx}\)

\(\large\color{slate}{\displaystyle\int\limits_{-1}^{0}8(x^5+3x)^3(5x^4+3)dx}\)

\(\large\color{slate}{8\displaystyle\int\limits_{-1}^{0}(x^5+3x)^3(5x^4+3)dx}\) (took the constant out)

Answer 7

then set: \(\large\color{slate}{u=x^5+3x}\)

Answer 8

when you get the answer tell me what you got.
you can change the bounds and work it, without having to substitute back for u,
or you can do it with u without changing the bounds, but not to forget to sub in the x back (which is more typical to do)

Answer 9

@SolomonZelman

Answer 10

\(\large\color{slate}{\displaystyle\int\limits_{-1}^{0}(x^5+3x)^3(5x^4+3)~dx}\)

you know how u-substitution works?

Answer 11

Well, if I were to do this problem: \(\large\color{slate}{\displaystyle\int\limits_{}^{}x^3(3x^2)~dx}\)

then you set: \(\large\color{slate}{u=x^3}\)
\(\large\color{slate}{du=3x^2~dx}\)

which would give you: \(\large\color{slate}{\displaystyle\int\limits_{}^{}u~du}\)
and from there it is easy.

Answer 12

I substituted some inner function instead of u,
||
V
(in my example's case it was x^3 )


and du consists of `derivative of u times dx`
||
V
(in my example's case it was 3x^2

Answer 13

did this make sense just now?

Answer 14

wait that was just an example? not from this problem right?

Answer 15

yes it was my own just now made up example

Answer 16

so do you understand the example, or not?

Answer 17

ohh ok cause i got confused haha, can we apply it together , and a little

Answer 18

in your problem, \(\large\color{slate}{\displaystyle\int\limits_{-1}^{0}(x^5+3x)^3(5x^4+3)~dx}\)

you are doing the same exact thing, as in my example.

I am setting \(\large\color{slate}{u=x^5+3x}\)

what is the derivative of u, can you tell me?

Answer 19

(I am asking for the derivative of \(\large\color{slate}{x^5+3x}\) )

Answer 20

5x^4+3

Answer 21

yes.

Answer 22

your \(\large\color{slate}{du}\) will (therefore) be

\(\large\color{slate}{du=(5x^4+3)dx}\)

Answer 23

excuse me I took a constant out, so the 8 should be there
like this, \(\large\color{slate}{8\displaystyle\int\limits_{-1}^{0}(x^5+3x)^3(5x^4+3)~dx}\)

Answer 24

So what I am doing here is this:

\(\large\color{blue}{8\displaystyle\int\limits_{-1}^{0}(x^5+3x)^3(5x^4+3)~dx}\)

\(\large\color{royalblue}{u=x^5+3x}\)

\(\large\color{royalblue}{du=(5x^4+3)~dx}\)

Answer 25

see how I am multiplying the dx times the derivative of u?

Answer 26

yes!

Answer 27

okay, so lets label the substitution:

\(\large\color{blue}{8\displaystyle\int\limits_{-1}^{0}(\color{red}{x^5+3x})^3\color{green}{(5x^4+3)~dx}}\)


\(\large\color{red}{u=x^5+3x}\) \(\large\color{green}{du=(5x^4+3)dx}\)


\(\large\color{blue}{8\displaystyle\int\limits_{-1}^{0}(\color{red}{u})^3\color{green}{du}}\)

Answer 28

see what I did?

Answer 29

yes!your explanation is great!

Answer 30

okay, can you integrate this (new) integral?

Answer 31

(just use the power rule)

Answer 32

oh, those are just the sums, we will work on how they apply here.

for now, imagine that you need to integrate an indefinite integral,

\(\large\color{blue}{8\displaystyle\int\limits_{}^{}(\color{red}{u})^3\color{green}{du}}\)

Answer 33

how would you integrate the above?

Answer 34

no need for the 2 C's, not even for one

Answer 35

\(\large\color{blue}{8\displaystyle\int\limits_{}^{}(\color{red}{u})^3\color{green}{du}}\)

\(\large\color{blue}{2\color{red}{u}^4+C}\)

Answer 36

that would be your result, and +C would be always on the outside...

Answer 37

you were correct, but you didn't reduce it

Answer 38

if you don't understand what I am trying to say, it's fine, I can explain more..

Answer 39

ohh

Answer 40

Oh, you understand,. i c... :) good

Answer 41

yeah i got it

Answer 42

got disconnected, dang it!

Answer 43

\(\large\color{blue}{8\displaystyle\int\limits_{~}^{~}u^3du=2u^4+C}\)

Answer 44

but we had a substitution that:

\(\large\color{blue}{u=x^5+3x}\) , right?

Answer 45

yes!

Answer 46

so substitute it back, into the result.

Answer 47

I mean substitute \(\large\color{blue}{x^5+3x}\) into the answer, instead of u.

Answer 48

you don't need to expand it, just tell me what you get after substitution

Answer 49

:(((((((((

Answer 50

\(\large\color{blue}{2u^4+C~~~~~~\Rightarrow~~~~2(x^5+3x)^4 }\) good?

Answer 51

cool

Answer 52

So this is what is going on here:
\(\large\color{blue}{\displaystyle\int\limits_{0}^{-1}(2x^5+6x)^3(5x^4+3x)dx}\)

\(\large\color{blue}{\displaystyle\int\limits_{0}^{-1}(2(x^5+3x))^3(5x^4+3x)dx}\)

\(\large\color{blue}{\displaystyle\int\limits_{0}^{-1}(2^3)(x^5+3x)^3(5x^4+3x)dx}\)

Answer 53

I will posting bit by bit, for you to process it

Answer 54

and each time, I am posting, I will start from the last thing I posted previous time

Answer 55

\(\large\color{blue}{\displaystyle\int\limits_{0}^{-1}(2^3)(x^5+3x)^3(5x^4+3x)dx}\)

\(\large\color{blue}{\displaystyle\int\limits_{0}^{-1}8(x^5+3x)^3(5x^4+3x)dx}\)

\(\large\color{blue}{8\displaystyle\int\limits_{0}^{-1}(x^5+3x)^3(5x^4+3x)dx}\)

Answer 56

Oh sorry... all steps are correct, except that I misplaced the 0 and -1 on top and bottom....

Answer 57

if you want me to slow down, say so...

Answer 58

So going from this: \(\large\color{blue}{8\displaystyle\int\limits_{-1}^{0}(x^5+3x)^3(5x^4+3x)dx}\)

Answer 59

please tell me we are almost done hahaha

Answer 60

\(\large\color{blue}{8\displaystyle\int\limits_{-1}^{0}(\color{red}{x^5+3x})^3\color{green}{(5x^4+3x)dx}}\)

\(\large\color{blue}{u=\color{red}{x^5+3x}~~~~~~~~~~~~~~~~du=\color{green}{(5x^4+3x)dx}}\)

\(\large\color{blue}{8\displaystyle\int\limits_{-1}^{0}(\color{red}{u})^3\color{green}{du}}\)

Answer 61

got it(:

Answer 62

Remembering that I multiply the \(\large\color{blue}{dx}\) times the derivative of \(\large\color{blue}{u}\) ?

So far so good?

Answer 63

okay, my replies in a thread got mixed up, so I went into my profl

Answer 64

got it

Answer 65

now,

\(\large\color{blue}{8\displaystyle\int\limits_{-1}^{0}(\color{red}{u})^3\color{green}{du}}\)


\(\large\color{blue}{2\color{red}{u}^4+C~\huge{|}{\Huge _{\color{white}{\frac{}{\color{blue}{-1}}}}^{\color{white}{\frac{}{\color{blue}{0}}}}}}\)

Answer 66

means it is an antiderivative from 0 to -1.