Math Is Fun Forum

glenn101

Member

Registered: 2008-04-02

Posts: 108

Integration

Need help with some integrals that I can't work out. Please don't use integration by parts, I have not learned it yet and it is not required to work out this integrals, use u-substitution.
1.

∫    e^2x
   ------------ dx
      1+e^x

2.
           e^x
∫ ------------------- dx
     e^2x-2e^x+1

3.
          If y=x√(4-x) find dy/dx and simplify. Hence evaluation : ∫0to2 8-3x/√(4-x) dx

4.
         The derivative of xlogex is 1+logex. An antiderivative of logex is equal to:

Please show complete working out for 3 & 4 especially, it's important that I know how to do this, my teacher never taught it to us, and I am on holidays at the moment so I can't contact him.

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"If your going through hell, keep going."

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Integration

Hi glenn101 ;

#1

Use these substitutions:

Last edited by bobbym (2009-07-01 17:44:57)

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

glenn101

Member

Registered: 2008-04-02

Posts: 108

Re: Integration

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.

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"If your going through hell, keep going."

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Integration

In the numerator:

u = u + (1 - 1) = (u + 1) - 1

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

glenn101

Member

Registered: 2008-04-02

Posts: 108

Re: Integration

Thanks for pointing that out Ricky,
any ideas for 2,3,4? it is essential for me to understand 3&4 especially.

---

"If your going through hell, keep going."

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Integration

Hi glenn101;

Sorry, for being cryptic on #1, I will try to be clearer next time.

#2

Same substitutions:

Now do these substitutions:

Now just do all your sustitutions:

So

Ricky and I were obviously posting at about the same time. I have done #2 for you, he has given you valuable insight into #3 and #4. Also he may have a niftier substitution for #2 then my double substitution of u and v, Try to work #2, #3, and #4 for yourself. If you get stuck come back here. Remember you have to do math to get good at math. Do lots of problems!

Last edited by bobbym (2009-07-02 19:10:44)

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Integration

For 2, remember that you don't want to sub u into the numerator.  Instead what do you want to sub?

For 3, you're going to have to be specific on what you're having trouble on.  Product rule is all you really need.

For 4, remember the fundamental theorem of calculus...  You want the derivative of some function to be log_e(x).  You know the derivative of x*log_e(x) is log_e(x) + 1.  That's close, but we need to get rid of the +1.  How do we do that?

---

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

glenn101

Member

Registered: 2008-04-02

Posts: 108

Re: Integration

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!

Last edited by glenn101 (2009-07-03 14:23:00)

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"If your going through hell, keep going."

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Integration

Now I get lost, what is it I do on this step?

Finish with the fundamental theorem of calculus.

---

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Integration

∫1+logex = xlogex
∫logex = xlogex-1

This is close, but not quite right.  Work out the transition from the first step to the second step much slower.  You'll see it.

---

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

glenn101

Member

Registered: 2008-04-02

Posts: 108

Re: Integration

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

Last edited by glenn101 (2009-07-03 16:10:04)

---

"If your going through hell, keep going."

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Integration

Hi glenn101;

If you have determined

and you want

Just observe that

The right hand side is twice the left hand side.
so

And by the way I noticed you are working on substitutions on the other thread. Excellent!

Last edited by bobbym (2009-07-03 18:08:27)

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

glenn101

Member

Registered: 2008-04-02

Posts: 108

Re: Integration

Ah, I see it now, thank for showing that.
And I'm glad you noticed me trying some of those substitutions, I still need the practise, so I thought I'd see if there are any on this site. And don't worry I've been thoroughly practising these with my textbook too.

---

"If your going through hell, keep going."

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Integration

Hi glenn101;

Excellent, by tomorrow you will be helping me.

Last edited by bobbym (2009-07-03 18:51:28)

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

glenn101

Member

Registered: 2008-04-02

Posts: 108

Re: Integration

Haha, I might:P
I also did Volumes of solids of revolutions before, fascinating stuff there.
I'm doing Differential Equations tomorrow

Last edited by glenn101 (2009-07-03 19:11:15)

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"If your going through hell, keep going."