Help with Calculus...

[JAZN 5,000+ posts]

that thing is ridiculously big. thats what she said! ohhhh

 

[swimfreak26 10+ year member]

lol, I'm sure I would love one too, if we were allowed to even use one

x2...not allowed to use calculators in any of my math courses in college. Calc 3 was a real kick in the *** when I depended on my calculator for calc 1 and 2 in high school, then all of the sudden I wasn't allowed to use it anymore.

 

[vettelovr1124 10+ year member]

The Voyage 200 is not acceptable for many tests...especially standardized tests.

I get to start Calc 3 my freshman year...sounds fun. At least I won't even touch anything new until a few weeks or months into the course. //content.invisioncic.com/y282845/emoticons/smile.gif.1ebc41e1811405b213edfc4622c41e27.gif

 

[JAZN 5,000+ posts]

whats calc 3? is that multivariable calc?

 

[electronicsgeek 10+ year member]

can someone help me with this problem...where did the 1/2 come from?

 

[ANeonRider 10+ year member]

Since du = 1/x, and you have 1/2x in the problem, you pull out the 1/2 so that the problem is 1/x... to match the du.

 

[electronicsgeek 10+ year member]

Since du = 1/x, and you have 1/2x in the problem, you pull out the 1/2 so that the problem is 1/x... to match the du.

not exactly understanding...i dont think 1/2 was in the problem...you get it somehow...where do you get the 1/2?

 

[ANeonRider 10+ year member]

look at the problem. It is (ln4x*1/2x)dx. Now when you find du, it equals 1/xdx. DU MUST EQUAL SOMETHING IN THE PROBLEM. So... since you have (1/2x)dx in the problem, you can pull out 1/2 so that it equals (1/x)dx, then you can continue the problem.

 

[electronicsgeek 10+ year member]

how would you do this problem? can you take me through step by step if its not to much work...

 

[BASS OUTLAW 10+ year member]

no shit .. i lost all my calculas maths skillz //content.invisioncic.com/y282845/emoticons/frown.gif.a3531fa0534503350665a1e957861287.gif

 

[sumone 10+ year member]

whenever you do u-substitution, you want to get rid of all x's and dx. You have to figure out what can "u" equal so that when the substitutions are made, all you end up with is u's & du.

In this problem,

u=ln 4x

du =       4
       ------ dx
         4x

du = x^-1 dx

dx =         du
       ------------
            x^-1
dx = x du 


pulling the 1/2 outside the integral and replacing ln 4x, with "u" and dx with "xdu" gives:
      _
1   |       u * xdu
---  |    --------------
2   |         x
    ¯ 

the x's cancel so that you're left with the integral of u du,
which is
               u²
            ------
               2

multiplying that times the 1/2 gives you
1
--- u²
4

putting ln 4x back in for u gives you the final answer, 

1
--- (ln 4x)²
4

 

[sumone 10+ year member]

and if you're confused where the 1/2 came from...it's part of the problem. you have ln 4x divided by 2x. you divide by two is same as multiply by 1/2. and you know that a constant (a normal number) can be pulled to the left side of the integral, hence 1/2 times the integral of ln 4x divided by x.

 

[electronicsgeek 10+ year member]

but why is 1/2 pulled out? to cancel something out?

 

[ANeonRider 10+ year member]

So that the du = 1/2x in the original equation.

 

[ANeonRider 10+ year member]

1/2 * 1/x = 1/2x basic freakin alegbra.