Question

@solomonzelman

Answer 1

What is exactly the problem here?

Continuing?
The expression for du?

Answer 2

Do I just sub u for 7x -2, then do the antiderivative, and then plug 7x - 2 back in?

Answer 3

You need an expression for du. (some expression that replaces dx)

Answer 4

Right?

Answer 5

\(\color{#000000 }{ \displaystyle \frac{ d }{dx } {\rm G}\left(f(x)\right)={\rm G}'\left(f(x)\right)~\times~f'(x) }\) (via the chain rule)

Therefore,

\(\color{#000000 }{ \displaystyle \int {\rm G}'\left(f(x)\right)~\times~f'(x)~dx}\)

\(\color{#000000 }{ \displaystyle u=f(x)}\)

\(\color{#000000 }{ \displaystyle \frac{du}{dx}=f'(x)\quad \Longrightarrow \quad du = f'(x)~dx }\)

and that substitution yields,

\(\color{#000000 }{ \displaystyle \int {\rm G}'\left(u\right)~du= {\rm G}\left(u\right)\color{grey}{\rm +C}}\)

and we substitute the original variable back,


\(\color{#000000 }{ \displaystyle \int {\rm G}'\left(u\right)~du= {\rm G}\left(f(x)\right)\color{grey}{\rm +C}}\)

Answer 6

You can say that we are kind of "applying the chain rule" when we integrate with u-substitution.

Answer 7

That is just conceptually...
so in your case,

\(\color{#000000 }{ \displaystyle 28\int (7x-2)^3~dx}\)

you are going to set,
\(\color{#000000 }{ \displaystyle u=7x-2}\)

\(\color{#000000 }{ \displaystyle \frac{du}{dx}=7\quad \Rightarrow \quad du=7~dx\quad \Rightarrow \quad \frac{1}{7}du=dx}\)

\(\color{#000000 }{ \displaystyle 28\int u^3~\cdot \left(\frac{1}{7}~du\right)\quad \Rightarrow \quad 4\int u^3 ~du}\)

Answer 8

In other words, you aren't just substitution u=something, and integrate.
You need to find an equivalent substitution for "dx"

Answer 9

AH! I see. Thank you. :)

Answer 10

I forgot about the du/dx aspect of the problem

Answer 11

yes, you always need that. (well, if you want to get the correct answer)

Answer 12

:D

Answer 13

lol, thx again. :)

Answer 14

Anytime ... also I have recently posted to another thread, where I played more with the u-substitution, and solved other components in the integral;
so if you want to see it at some time, I'll find it an post it below....

yw

Answer 15

Okay, I will look at that.

Answer 16

there,

http://openstudy.com/users/friedrice14#/updates/568c6030e4b0b72ba793f909

Answer 17

Anytime:) ...