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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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 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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 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. Tags antiderivative, exponent, find Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Shamieh Calculus 1 November 3rd, 2013 06:59 PM mortified_penguin Calculus 7 January 24th, 2013 08:03 AM Valar30 Calculus 2 November 15th, 2010 11:58 AM Valar30 Calculus 1 November 15th, 2010 09:03 AM lovetolearn Algebra 1 December 31st, 1969 04:00 PM

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