Question

Integral sqrt(4+3x) dx

Answer 1

\[\int\limits_{}^{} \sqrt{4+3x} dx\]

Answer 2

let 4 + 3x = u
then 3dx = du

or, \( 1/3 \int \sqrt{u} du\)
= \( 1/3 u^{3/2}/(3/2) \)
= \( 2/9 * (4+3x)^{3/2} \)

Answer 3

can someone do it without u substitution if possible i struggle with that

Answer 4

\[2(4+3x)^{3/2}\]

Answer 5

I got my result by first applying the standard rule for integration with respect to (4+3x).Then I differentiated the result with respect to x and found the result was twice the required result. So my result should have been:
\[1/2(4+3x)^{3/2}\]

Answer 6

wouldnt the standard rule just lead you to (2/3)(4+3x)^(3/2)?

Answer 7

Correct

Answer 8

@Redlinl , you might want to practice u substitution, because i'm pretty sure you'll see more of it.

Answer 9

\[\int\limits \left( ax + b \right)^n dx = \frac{1}{a \left( n + 1 \right)}\left( ax + b \right)^{n + 1} + C\]\[\int\limits \sqrt{4 + 3x} dx = \int\limits \left( 3x + 4 \right)^{\frac{1}{2}} dx\]\[= \frac{1}{3\left( \frac{1}{2} + 1 \right)}\left( 3x + 4 \right)^{\frac{1}{2} + 1} + C\]\[= \frac{2}{9} \left( 3x + 4 \right)^{\frac{3}{2}} + C\]\[= \frac{2}{9}\left( 3x + 4 \right)\sqrt{3x + 4} + C\]

Answer 10

@redlinl You're gonna need U-substitution for calculus later on. U-Sub is actually the simplest technique for integration compared to the stuff you learn later on.

Answer 11

okay but exraven will i need to memorize that property or is there any way to prove that easily?

Answer 12

@Redlinl , that's the chain rule.

Answer 13

well, more like reverse chain rule of differentiation.

Answer 14

you dont need to memorize it, u-substitution is the best way

Answer 15

Then when you differentiate back you get:\[(4+3x)^{1/2}\times3=3\sqrt{(4+3x})\]

Answer 16

could i just integrate it in the way i would to get the int by parts formula with the chain rule?

Answer 17

By the way, anyone wanna provide me with a site where I can learn u-substitution? I can't do it for life.

Answer 18

Red, just learn integration with u-sub

Answer 19

You aren't going to pass the course without it.

Answer 20

Which is three times the original expression to be integrated.

Answer 21

well im not in any course now im just preparing for it by learning from a book but i could also use a good site

Answer 22

here you go http://tutorial.math.lamar.edu/Classes/CalcI/SubstitutionRuleIndefinite.aspx

Answer 23

Well, I can too, because I'm learning it from a book also, and it's an 80s textbook ;)

Answer 24

I think differentials are not explained very well.

Answer 25

The result is that I can't use them.

Answer 26

yeah and what does du actually mean? d/dx u?

Answer 27

So the correct result is as follows:
\[2/9(4+3x)^{3/2}\]
Sorry about my earlier errors:(

Answer 28

You can manipulate differentials like algebraic terms. (du) just means an infintely small change in (u)

Answer 29

so the du is the derivative of u?

Answer 30

the way i think of it, du means delta u, and dx means delta x, so dx/du, means delta x/ delta u, which is just the slope or the derivative. Remember, when you're integrating, you're actually approximating the area of the curve by adding an infinite number of boxes and their area. The dx and du is the base of the box, multiplied by the f(x) which is their height, you get the area of the boxes.

Answer 31

does the d actually stand for \[\Delta\] because that would make this a little easier

Answer 32

Yea thats how i think of them. In the Reimann sum we had the capital Delta, and in the integral we had dx, so they basically both mean the base of the rectangles.

If you ever get to Arc Length of a curve, they'll explain a little more about what dy/dx really means.

Answer 33

\[\sqrt[3]{?3x-1=3}\]