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 May 8th, 2016, 05:10 AM #1 Member   Joined: Apr 2016 From: Hell Posts: 39 Thanks: 0 How to find the exponent of antiderivative? Imgur: The most awesome images on the Internet I don't understand the explanations in the text. How would this problem be solved? Also, what is the term for the stylized f?
 May 8th, 2016, 05:16 AM #2 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,838 Thanks: 653 Math Focus: Yet to find out. The stylized f? You mean the integral sign? $\displaystyle \int$ Do you know what an antiderivative is? Thanks from Incomprehensible
May 8th, 2016, 05:22 AM   #3
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Quote:
 Originally Posted by Joppy The stylized f? You mean the integral sign? $\displaystyle \int$
Yes.
Quote:
 Do you know what an antiderivative is?
That's the opposite of a derivative? I don't understand how the math translates from one to the other.

 May 8th, 2016, 05:32 AM #4 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,838 Thanks: 653 Math Focus: Yet to find out. As a simple example, consider $\displaystyle y = x^2$ The derivative would be, $\displaystyle \frac{dy}{dx} = 2x^{2-1}$ Conversely, $\displaystyle \int{2x} dx = \frac{2x^{1+1}}{2} + C$ But I'm guessing this won't help you, as there's quite a bit going on behind the scenes. As I've mentioned before, this isn't a pattern matching game, you have to understand it first. Thanks from Incomprehensible Last edited by skipjack; May 8th, 2016 at 09:10 AM.
May 8th, 2016, 05:44 AM   #5
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Quote:
 Originally Posted by Joppy Conversely, $\displaystyle \int{2x} dx = \frac{2x^{1+1}}{2} + C$
So, the process is to take the 2x, add one to the power, then divide it by two? And then add a C?

Is this consistent for all problems? Also, what is the C for?

 May 8th, 2016, 05:49 AM #6 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,838 Thanks: 653 Math Focus: Yet to find out. I should have expressed it in general form, sorry. That way you can see what the definition is. I'm not on a device capable of using latex currently (at least not conveniently), so I will redirect you to a website. www.mathsisfun.com/calculus/integration-introduction.html I think this is a reasonable, and fairly simple introduction. I suggest that you read through it, and take some notes. Thanks from Incomprehensible Last edited by skipjack; May 8th, 2016 at 09:17 AM.
 May 9th, 2016, 11:56 PM #7 Member   Joined: Apr 2016 From: Hell Posts: 39 Thanks: 0 How would anti differentiating work for negative exponents? $6x^{-2} + 2x^{-4} -3x^{-3}$. How would the anti diff process work for fractions? I have $\frac{1}{6x^2}$ for the first part. Am I supposed to add one to x, and then put it all over 3? How would the fraction then resolve?
 May 10th, 2016, 12:30 AM #8 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,838 Thanks: 653 Math Focus: Yet to find out. It works the same way. $\displaystyle \int{x^n}{dx} = \frac{1}{n+1} x^{n+1}$ But be careful of the case when you have $\displaystyle x^{-1}$ $\displaystyle \int\frac{1}{x}{dx} = \int{x^{-1}}{dx} = ln|x|$ Does this help? Also note that.. $\displaystyle \frac{1}{6x^2} â‰ 6x^{-2}$ Thanks from Incomprehensible Last edited by Joppy; May 10th, 2016 at 12:36 AM.
May 10th, 2016, 01:47 AM   #9
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Quote:
 Originally Posted by Joppy It works the same way. $\displaystyle \int{x^n}{dx} = \frac{1}{n+1} x^{n+1}$ But be careful of the case when you have $\displaystyle x^{-1}$ $\displaystyle \int\frac{1}{x}{dx} = \int{x^{-1}}{dx} = ln|x|$ Does this help? Also note that.. $\displaystyle \frac{1}{6x^2} â‰ 6x^{-2}$
So, it's 6 / x^2? Then add one to x for x^-1, then dividing by that same -1, for -6x^-1?

May 10th, 2016, 03:47 AM   #10
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Quote:
 Originally Posted by Incomprehensible So, it's 6 / x^2? Then add one to x for x^-1, then dividing by that same -1, for -6x^-1?
Ok let's try work through it, see this rule here,

$\displaystyle \int{x^n}dx = \frac{1}{n+1}x^{n+1}$

Let's apply this to your problem.

$\displaystyle \int{\frac{x^{-2}}{6}}dx$ (we can take constants outside of the integral sign here, so lets do that to tidy things up).

$\displaystyle \frac{1}{6}\int{x^{-2}}dx$

Now look at the rule i've provided at the top, can you identify n?

It's -2 would you agree? That is, n = -2. Now let's substitute this back into the expression above, we have,

$\displaystyle \int{x^n}dx = \frac{1}{n+1}x^{n+1}$ now substitute n, $\displaystyle \frac{1}{6}\int{x^{-2}}dx = \frac{1}{-2+1}x^{-2+1} = -\frac{1}{6x}$

Do you think you could apply a similar procedure to a different problem? Have a go integrating 2x^-4

Last edited by Joppy; May 10th, 2016 at 04:10 AM.

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