Math Is Fun Forum

Ah but Ricky,
at that point I'm evaluating the integral
            8-3x
             ---------
             2√(4-x)
the question requires the integral of;

            8-3x
          ----------
            √(4-x)
How do I get to that step?
oh and for #3

x+∫logex= xlogex
hence ∫logex=xlogex-x

Sorry about not clearly stating what I'm having trouble with Ricky/bobbym
I will now try and do questions 3 & 4 here.

y=x√(4-x)
Product Rule:
Let u =x, v=√(4-x)
du/dx=1, dv/dx=     -1
                         ----------
                           2√(4-x)
dy/dx= u(dv/dx)+v(du/dx)
         =     -x
             -----------+ √(4-x)       
              2√(4-x)
                -x              2(4-x)     
         =   -----------+  ---------
              2√(4-x)      2√(4-x)
              2(4-x)-x
         =   ----------
               2√(4-x)
         =      8-3x
               ---------
                2√(4-x)
            ok there's the derivative, I'm confident thats correct.
Now this is where I get lost.
             
            ∫(0 to 2) 8-3x
                        --------- = [x√(4-x)](0 to 2)
                         2√4-x
Now I get lost, what is it I do on this step? my surd combined with fractions algebra is horrific.

now for #4.

∫1+logex = xlogex
∫logex = xlogex-1
ah, I think I get that one now, can't believe I missed that.
And it's so true bobbym about practise, it's just with these I continuously get wrong so I'm practising my mistakes, once I've fixed these up I'm going to practise Heaps!:D
Oh and thanks for showing excellent working out for #2, I never would have thought of using 2 substitutions!

Thanks for the quick reply Bobbym,
I follow you untill you get to this step:

how does the left hand side = the right hand side? how did you seperate the integrals? I havn't seen that before.

ok for your first 1:
First thing you want to do is express to same base:
4^(x-5)=16^x
4^(x-5)=4^2*x
hence now we can evaluate powers;
x-5=2x
-5=x
x=-5
Remember, when working with indices or logs, you always have to have the same base to start evaluating.

Ok for 2:
I'm assuming those logs are base 10;
log3x=log18
take log18 to other side
log(x/6)=0
x=6 as log1=0.
Hope this helps:D

Ah, so either works. Thanks for verifying that. I think I best get on practicing these problems. I can't thank you enough for your efforts bobbym.

Thanks for the additional information bobbym.
I think your form works under the conditions you stated, mine presented in my textbook is as follows:
                a
∫      ------------        = Tan^-1(x/a)+c
         a^2+x^2

hmm I see what you mean, I just can't find a definate way of getting rid of that 4 on the denominator, thats what I'm trying to achieve. I'm getting completely confused now, I've been at this the whole day:(

I'll show you another question which I used the same approach, this one worked however.
          1
∫ ----------------
     9+16t^2
             1
= ∫ ------------------
        16(9/16+t^2)
           3/4
= 4/3∫ ----------------
           16(9/16+t^2
                   12
= 1/12∫ --------------
              16(9/16+t^2)
                    3
= 1/12∫ ----------------
              4(9/16+t^2)
                     3/4
= 1/12∫-----------------
             9/16+t^2

= 1/12Tan^-1(4/3t)+c

Could someone show me how they would go about integrating this:|

lol! good one Devante

That is an interesting pattern indeed!, will definatly help, requires minimal memory!
thanks for sharing that mathsy!

ah ok fair enough, using a lot of memory isn't good at all though.
I think I'll do what your doing, and stick with good ol' Firefox.