integration (1 Viewer)

[lanvins]

integrate

1. 6/(5-2x) dx

2. 6(1-2x)^-2 dx

3. (4-3x)^-1 dx

4. find y in terms of x for each of the following
dy/dx= 1/(2x) and y=2 when x=e^2

Thanks

 

[12o9]

 

[lolokay]

integrate

1. 6/(5-2x) dx

2. 6(1-2x)^-2 dx

3. (4-3x)^-1 dx

4. find y in terms of x for each of the following
dy/dx= 1/(2x) and y=2 when x=e^2

Thanks

y' = 6/(5-2x) dx
= -3(-2/(5-2x))
y = -3log(5-2x) +C

2. y' = 6(1-2x)^-2 dx
let u = (1-2x)
y' = 6u^-2.du.(dx/du)
dx/du = -1/2
y = -1/2*-6u^-1 +C
y = 3(1-2x)^-1 +C

3. y' = (4-3x)^-1 dx
= -1/3 -3/(4-3x).dx
y = -1/3 log(4-3x) +C

4. dy/dx= 1/(2x) and y=2 when x=e^2
y' = 1/2*2/2x => y = 1/2 log 2x +C
2 = 1/2*log 2e^2 +C
= 1/2*(2 + log2) + C
C = 1 - 1/2*log2
y = 1/2 log 2x + [1 - 1/2*log2]

 

[lanvins]

For

Q1, the book has it as -3loge |2x-5|+c, where the 2x is positve and the 5 is negative. why is that?

Q2. i don't get the point of this step
y' = 6u^-2.du.(dx/du)
dx/du = -1/2

Q3. same problem as question 1, the book has it as y = -1/3 log|-4+3x| +C

Q4. the back of the said the answer is 1/2loge |x|+1

Thanks for helping me by the way

 

[Trebla]

The absolute value accounts for negative values in the logarithm as well. For example if we integrate 1/x we get ln x + c; but ln x is only defined for x > 0. We know that the curve y = 1/x can be defined for negative x as well, so we use absolute value in the logarithms as ln |x| to help define the negative values of x.

I think you should be alright without absolute values in the logarithms. It might be a bit confusing for ext1. They are still correct with or without them, at least within the context of the HSC course.
Just remember if you ever have a sum or difference of two logs, it is best to combine them as one log immediately to avoid confusion.
e.g. integrate y = 1/x from x = -2 to x = -1 gives
[ln x]^-2_-1
= ln (-2/-1) => immediately combine the difference of two logs or else we get logs of negative numbers (alternatively you could use absolute value)
= ln 2

 

[lolokay]

For

Q1, the book has it as -3loge |2x-5|+c, where the 2x is positve and the 5 is negative. why is that?

Q2. i don't get the point of this step
y' = 6u^-2.du.(dx/du)
dx/du = -1/2

Q3. same problem as question 1, the book has it as y = -1/3 log|-4+3x| +C

Q4. the back of the said the answer is 1/2loge |x|+1

Thanks for helping me by the way

q1,3 what Trebla said ^

q2. you don't really need to do it like that. It's just using substitution and evaluating dx/du so it will be in a form you can inegrate

q4. the book is right. I forgot to combine logs.

 

[vivid]

For

Q1, the book has it as -3loge |2x-5|+c, where the 2x is positve and the 5 is negative. why is that?

Q2. i don't get the point of this step
y' = 6u^-2.du.(dx/du)
dx/du = -1/2

Q3. same problem as question 1, the book has it as y = -1/3 log|-4+3x| +C

Q4. the back of the said the answer is 1/2loge |x|+1

Thanks for helping me by the way

Question 4: The problem we have here is our basic log principles.

Everyone has agreed so far that y=1/2ln2x+c, yes?

Sub in the point (e^2, 2)

2=1/2(ln2e^2)+c
Let us remember that this is the logarithm of 2(e^2)
Thus, we cannot turn it into 2ln2e.
We can only do that if we it is ln(2e)^2, agreed?

So:
1/2(ln(2xe^2)+c=2
1/2(ln2+lne^2)+c=2
1/2(ln2+2lne)+c=2
1/2(ln2+2x1)+c=2
1/2ln2+1/2x2+c=2
1/2ln2+c+1=2
c=1-1/2ln2

Sub the value of c back in to get:

y=1/2ln2x+(1-1/2ln2)
y=1/2(ln2x-ln2)+1
y=1/2(ln(2x/2))+1
y=1/2lnx+1

It has already been explained that the absolute value is regarding the doman of the function. Just be careful with logs - make sure that in ln(f(x)^n), the power is for the whole of f(x) before you move it to the front.

Oh and in regards to question 2, there is the 'reverse chain rule' which is not really a reverse chain rule as it only works for linear functions. i.e. when integrating f(x)^n, f(x) must be linear (the greatest power of x is 1). I can't remember this rule right now, but yeah it should be in your textbook.