Antiderivatives Part II

[Hckyplayer8]

The indefinite integral of (x^1/2-3x^11/17+x^-2) = ?

 

[topsquark]

Try out that power law I mentioned in the other thread. For the first term we have the power law with n = 1/2. Give it a shot and post back here with the results.

-Dan

 

[Steven G]

Please write dx at the end of the integral.
So what is the problem, Dan gave you the formula on your last post.
Here are some rules to solve this integral
1) int[ f(x) +/- g(x)]dx = int f(x)dx +/- int g(x) dx. That is integrate each term separately.
2) int x^n dx = x^(n+1)/(n+1) + c
3) Here is a trick to adding 1 to a fraction. a/b + 1 = a/b + b/b = (a+b)/b. That is add 1 to a/b, the numerator will be (a+b) and the denominator will be b. So 4/11 + 1 = (4+11)/11 or 15/11

 

[Hckyplayer8]

Based off my previous thread, I get the basic idea of what we are trying to do.

To me, this is a lot like factoring...which my skill set is pretty sub-par. So beyond just guessing numbers, I really don't know what to do here.

The x^1/2 looks pretty easy. Add n + 1. So that gets me 3/2. Multiplying that by 2/3 gets me the original 1x and completes the antiderivative of that particular term.

 

[Steven G]

Based off my previous thread, I get the basic idea of what we are trying to do.

To me, this is a lot like factoring...which my skill set is pretty sub-par. So beyond just guessing numbers, I really don't know what to do here.

The x^1/2 looks pretty easy. Add n + 1. So that gets me 3/2. Multiplying that by 2/3 gets me the original 1x and completes the antiderivative of that particular term.

Why factoring? You are adding 1 to get the new power and dividing by 1 more than the original power or multiply by the reciprocal of the 1 more than the numerator

 

[Hckyplayer8]

Try out that power law I mentioned in the other thread. For the first term we have the power law with n = 1/2. Give it a shot and post back here with the results.

-Dan

Please write dx at the end of the integral.
So what is the problem, Dan gave you the formula on your last post.
Here are some rules to solve this integral
1) int[ f(x) +/- g(x)]dx = int f(x)dx +/- int g(x) dx. That is integrate each term separately.
2) int x^n dx = x^(n+1)/(n+1) + c
3) Here is a trick to adding 1 to a fraction. a/b + 1 = a/b + b/b = (a+b)/b. That is add 1 to a/b, the numerator will be (a+b) and the denominator will be b. So 4/11 + 1 = (4+11)/11 or 15/11

Thanks for the quick replies! I posted my first attempt at x^1/2.

I have to write the dx to specify which variable I'm integrating?

Thank you for the tricks/tips.

 

[JeffM]

Derivatives and anti-derivatives are inverse operations, analogous to addition/subtraction or multiplication/division.

How do I know I did a subtraction correctly? I add what I subtracted to my answer to see whether it gets me back to where I started.

[MATH](a - b) + b = a - (b - b) = a - 0 = a.[/MATH]
How do I know I did a division correctly? I multiply my answer and what I divided the quotient and the divisor to see whether it gets me back to where I started.

[MATH]\dfrac{a}{b} * b = a * \dfrac{b}{b} = a * 1 = a.[/MATH]
So if I take the derivative of f(x)'s antiderivative, I get back to where I started (at least if I made no error).

[MATH]f(x) = x^{1/2} \implies \int x^{1/2} \ dx = \left ( \dfrac{1}{1 + \dfrac{1}{2}} * x^{\{1 + (1/2)\}} \right ) + C =[/MATH]
[MATH]\dfrac{1}{\dfrac{3}{2}} * x^{3/2}+ C = \dfrac{2}{3} x^{3/2} + C.[/MATH]
Is that correct?

Differentiate it to check.

[MATH]\dfrac{d}{dx} \left ( \dfrac{2}{3} x^{3/2} + C \right ) = \dfrac{3}{2} * \dfrac{2}{3} * x^{\{(3/2) - (1/2)\}} + 0 = x^{1/2}.[/MATH]
Right back where we started so it is indeed correct.

You never have to be uncertain whether you did an anti-derivative correctly.

And yes, the dx indicates that you are integrating with respect to x and also acts as a delimiter. It may seem silly to worry about what you are integrating with respect to now, but when you get to u-substitutions, it is critical.

 

[Steven G]

I have to write the dx to specify which variable I'm integrating?

Yes, always!

 

[pka]

I have to write the dx to specify which variable I'm integrating?

Yes, always!

That is totally false. There is absolutely no reason to require the \(\displaystyle dx\) when writing an indefinite integral.
In fact, I know many instructors (particularly those who were Wall"s students or Wall's decedents) would mark off points if the \(\displaystyle dx\) were included. So you must follow whatsoever your instructor requires.

Allow me to give you a sample of Wall's approach. \(\displaystyle \int {\sqrt x } =?\) asks what expression has as its derivative \(\displaystyle \sqrt x~?\)
What is the exponential form of \(\displaystyle \sqrt x~?\) Is it \(\displaystyle x^{\frac{1}{2}}~?\)
What might we differentiate to get \(\displaystyle x^{\frac{1}{2}}~?\) Is it \(\displaystyle x^{\frac{3}{2}}~?\)
But it is \(\displaystyle {\bf 1}x^{\frac{1}{2}}~?\) Is it \(\displaystyle {\tfrac{2}{3}}x^{\frac{3}{2}}~?\)

 

[HallsofIvy]

I don't know who "Wall" is but writing "\(\displaystyle \int f(x)\)" rather than "\(\displaystyle \int f(x)dx\)" seems completely wrong!

 

[topsquark]

As a Physicist I pretty much need the "dx." I have looked at some "forms" and note that they lack the dx, but I'm not really up on those yet.

As a pet peeve: [math]\int 3x - 2x^2 dx[/math] is an issue for me, though I've heard that many Mathematicians use it rather often.

-Dan

 

[Hckyplayer8]

Following the lead of the above the antiderivavtive of the first term is 2/3x^3/2

Now for the second term, first 3x^11/17+17/17 = 3x^28/17

Now I separate the constant and apply the power rule. [x^28/17] / [28/17] = [(17)(x^28/17) / 28]

Then multiply by the constant for a final term of 51/28 x ^{28 / 17}

 

[Hckyplayer8]

Lastly x^-2 = x^-1 / -1 (after the power rule) which equals -x^-1 which equals -1/x

Add on the constrant reveals the final answer is [2/3 x^3/2] - [51/28 x ^28/17] - 1/x + C

 

[JeffM]

As a Physicist I pretty much need the "dx." I have looked at some "forms" and note that they lack the dx, but I'm not really up on those yet.

As a pet peeve: [math]\int 3x - 2x^2 dx[/math] is an issue for me, though I've heard that many Mathematicians use it rather often.

-Dan

Dan

That's why I said it acts as a delimiter. Everything between the integral symbol and the differential is what is being integrated.

 

[Steven G]

Following the lead of the above the antiderivavtive of the first term is 2/3x^3/2

Now for the second term, first 3x^11/17+17/17 = 3x^28/17

Now I separate the constant and apply the power rule. [x^28/17] / [28/17] = [(17)(x^28/17) / 28]

Then multiply by the constant for a final term of 51/28 x ^{28 / 17}

I

Lastly x^-2 = x^-1 / -1 (after the power rule) which equals -x^-1 which equals -1/x

Add on the constrant reveals the final answer is [2/3 x^3/2] - [51/28 x ^28/17] - 1/x + C

Please do not write x^-2 = x^-1 / -1. You know they are not equal so why use the symbol that says that they are equal? You are in calculus now, so try to write things correctly.

 

[topsquark]

Dan

That's why I said it acts as a delimiter. Everything between the integral symbol and the differential is what is being integrated.

Oh I know that, and really it's fine. It's just that in a day when Math teachers, (and Physics teachers!), textbooks, and computer aided classes all think that they can use 3x +2/5x -3 to represent (3x + 2)/(5x - 3) I just get a little titchy about grouping.

-Dan