Question

Question deleted.

Answer 1

\[\Large \int \sqrt{3x-5}dx\]

Answer 2

Okay, that was weird, but anyway, if it were just the integral of the square root of something, then...it'd be easy, right?

Answer 3

If only it were something like this:
\[\Large \int \sqrt u \ du\]

Answer 4

@Phayeth do you copy? lol

Answer 5

yes, i do so far..

Answer 6

So, there must be a u-substitution that comes to mind...
What does your instinct say?

Usually, you'll let u be that thing inside the most complicated function in your integrand... in this case, it's the square root.

Try u = 3x - 5
and tell me what happens next...

Answer 7

you find the derivative of 3x-5?

Answer 8

Which is...?

Answer 9

3

Answer 10

Okay, so you have
u = 3x - 5
du = 3dx

So make the substitution, and tell me what your integral becomes...

Answer 11

\[\sqrt{3x-5}*3 dx ?\]

Answer 12

Nope. You cannot just make things magically appear just like that... there are rules involved...

\[\Large \int \sqrt{3x - 5} \ dx\]

Answer 13

We'll turn this into an integral along the variable u instead.
Can you express the integrand in terms of u?

Answer 14

\[\frac{ 2 }{ 3 }*u^\frac{ 3 }{ 2 }\] ?

Answer 15

LOL magic? Show how you got that...

Answer 16

Can i explain it?

you do the opposite of differentiation.
you raise the power by one and then divide by the reciprocal...

Answer 17

But why is the exponent of u 3/2?

Answer 18

oh, the square can also be expressed as to the power of a half.
so i added 1 to the 1/2 and i got 3/2.

Answer 19

I know the square ROOT is expressed as a half-power, but why did you add 1?

Answer 20

becuase its integration?

Answer 21

Okay... I'll demonstrate...
\[\Large \int \sqrt{3x - 5} \ dx\]

Answer 22

We let u = 3x - 5

So THIS part
\[\Large \int \sqrt{\color{red}{3x - 5}} \ dx\]
Could be replaced with u.
\[\Large \int \sqrt{\color{red}u} \ dx\]

Answer 23

But we need the differential to be du instead of dx.
How do we substitute?

Answer 24

du/dx = 3
Isolate du
it then becomes du=3*dx..

Answer 25

Okay, or equivalently,
\[\Large \frac{du}3 = dx\]
right?

Answer 26

ok.

Answer 27

no wait, im confused..

Answer 28

du = 3dx.
I just divided both sides by 3.

Answer 29

yes.

Answer 30

So we can replace dx, as follows
\[\Large \int \sqrt{u} \ \color{red}{dx} \qquad \rightarrow \qquad \int \sqrt u \ \color{blue}{\frac{du }{3}}\]

Answer 31

oh okay. i get you now...

Answer 32

Good.
Now...
\[\Large = \frac13 \int \sqrt u \ du\]
And the rest is history.

Answer 33

okay. Thank you. :)

Answer 34

No problem ^^