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May 8th, 2016, 05:10 AM   #1
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How to find the exponent of antiderivative?

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I don't understand the explanations in the text. How would this problem be solved? Also, what is the term for the stylized f?
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May 8th, 2016, 05:16 AM   #2
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The stylized f? You mean the integral sign? $\displaystyle \int$

Do you know what an antiderivative is?
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May 8th, 2016, 05:22 AM   #3
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Quote:
Originally Posted by Joppy View Post
The stylized f? You mean the integral sign? $\displaystyle \int$
Yes.
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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.
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May 8th, 2016, 05:32 AM   #4
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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.
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Last edited by skipjack; May 8th, 2016 at 09:10 AM.
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May 8th, 2016, 05:44 AM   #5
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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?
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May 8th, 2016, 05:49 AM   #6
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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.
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Last edited by skipjack; May 8th, 2016 at 09:17 AM.
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May 9th, 2016, 11:56 PM   #7
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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?
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May 10th, 2016, 12:30 AM   #8
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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.. $\displaystyle \frac{1}{6x^2} ≠ 6x^{-2}$
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Last edited by Joppy; May 10th, 2016 at 12:36 AM.
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May 10th, 2016, 01:47 AM   #9
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Quote:
Originally Posted by Joppy View Post
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.. $\displaystyle \frac{1}{6x^2} ≠ 6x^{-2}$
So, it's 6 / x^2? Then add one to x for x^-1, then dividing by that same -1, for -6x^-1?
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May 10th, 2016, 03:47 AM   #10
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Originally Posted by Incomprehensible View Post
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
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Last edited by Joppy; May 10th, 2016 at 04:10 AM.
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