Anti-Derivative

[CanadaAotS]

Hello all.

 

I just had a calculus question. Were learning all about derivatives and everything right now.

 

I was just wondering how do you get the anti-derivative of something? Like working backwards from the derivative.

 

I know that if

[math]

f'(x) = 2x

[/math]

then working backwards it'd be

[math]

f(x) = x^2 + c

[/math]

where c could be anything, but I dont know why.

 

If anyone could show me an equation of some sort that gives you an anti-derivative I'd be happy.

[cosine]

Hello all.

 

I just had a calculus question. Were learning all about derivatives and everything right now.

 

I was just wondering how do you get the anti-derivative of something? Like working backwards from the derivative.

 

I know that if

[math]

f'(x) = 2x

[/math]

then working backwards it'd be

[math]

f(x) = x^2 + c

[/math]

where c could be anything' date=' but I dont know why.

 

If anyone could show me an equation of some sort that gives you an anti-derivative I'd be happy.[/quote']

 

A method for finding anti-derivitives is to invert the differentiation rules. The simplest example is for polynomials.

[math]\frac{d(x^{n})}{dx} = nx^{n-1}[/math]

so given [math]ax^{b}[/math]

you get:

[math] y = \frac{ax^{b+1}}{b+1} [/math]

You will learn more rules soon. If you have a calculus textbook there is probably a reference inside the front or back cover. (Though of course I suggest you read the anti-deritive chapter).

 

And this is my first attempt with Latex.... hope it works

[Dave]

Typically we call the "anti-derivative" of a function the integral of a function. You can denote it like this:

 

[math]\int f(x) \, dx[/math]

 

There's no explicit formula for the integral, and in many cases you can't even integrate a function. You might find it a relatively fun exercise to determine a formula for:

 

[math]\int x^n \, dx[/math]

 

Where n is any number not equal to -1 (why?) I'm reticant to give too much away since you've just started differentiating, but you'll probably touch on it soon enough anyway.

[CanadaAotS]

We dont get calculus texts cause its a highschool course, but it would help if say the simple example I gave

[math]

f(x) = 2x

[/math]

and you find the anti-derivative (or integral I guess as its called) then I'd get anything else as the process would be shown.

 

Even though will probably be doing this in a week or 2, getting a general feel before hand would definetly be helpful.

[BobbyJoeCool]

Where n is any number not equal to -1 (why?) I'm reticant to give too much away since you've just started differentiating, but you'll probably touch on it soon enough anyway.

 

He he... does it have to do with x^0=1, and d1/dx=0 and x^-1 cannot be 0?

 

 

As you haven't dealt with intrigals before... just think of derivitive, but backwards...

 

since it's a simple f(x)=2x... to take the derivative of a function (power rule), you multiply by the exponent, and subtract one from the exponent... to take the "anti-derivative.." go backwards... add one to the exponent, then divide by the exponent...

 

so, general case...

 

[math]f(x)=ax^n[/math]

 

[math]\int ax^n \, dx=\tfrac{ax^{n+1}}{n+1}[/math]

 

But remember... the derivative of ANY constant is 0... so when taking the intrigal.. you have to add c, where c is any constant... so...

 

[math]\int ax^n \, dx=\tfrac{ax^{n+1}}{n+1}+c[/math]

 

the easy way to check to see if you did an intrigal correctly... take the derivative!!!

 

with your example...

 

[math]\int 2x^1 \, dx=\tfrac{2x^{1+1}}{1+1}+c[/math]

 

[math]\int 2x \, dx=\tfrac{2x^{2}}{2}+c[/math]

 

[math]\int 2x \, dx=x^{2}+c[/math]

 

Take the derivative to check your answer...

 

[math]x^2+c[/math]

 

[math]\tfrac{d}{dx}x^2+c[/math]

 

[math]2x+0[/math]

 

[Dave]

Also, the motivation behind integrating functions is often not to find the anti-derivative, but the useful fact that if you can integrate a function then you can find its exact area under the curve. For example,

 

[math]\int_{0}^{1} x^2 \, dx = \tfrac{1}{3}[/math].

 

It's a little tricky at first, but I daresay you'll get used to it after a while - much better than counting squares, and a lot faster!

[BobbyJoeCool]

that looks farmiliar, but... I can't remember if we did that in Intro to Calc in High School... can you give me a basic idea of how do get...

 

[math]\int_{0}^{1} x^2 \, dx = \tfrac{1}{3}[/math]

 

Just to see if it jogs a memory of it?

 

Much faster than counting squares (especially if you want an exact answer on a curve and hsve to use limits!)

[TD]

Thanks to the fundamental theorem of calculus, you can compute these (definite) integrals by using anti-derivatives. If F(x) is an anti-derivative of f(x), then:

 

[math]\int\limits_a^b {f\left( x \right)dx} = \left[ {F\left( x \right)} \right]_a^b = F\left( b \right) - F\left( a \right)[/math]

 

The middle step is just a notation and is not required.

Can you figure it out now for that example?

[CanadaAotS]

I get it, makes sense now thanks all!

[BobbyJoeCool]

Thanks to the fundamental theorem of calculus' date=' you can compute these (definite) integrals by using anti-derivatives. If F(x) is an anti-derivative of f(x), then:

 

[math']\int\limits_a^b {f\left( x \right)dx} = \left[ {F\left( x \right)} \right]_a^b = F\left( b \right) - F\left( a \right)[/math]

 

The middle step is just a notation and is not required.

Can you figure it out now for that example?

 

Yea, I remember now... this was like a week before graduation, and everyone had one foot out the door...

[TD]

Too bad, it's important ;o)