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May 11th, 2016, 02:53 AM   #11
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Quote:
Originally Posted by Joppy View Post
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
Is that -6x^-3

How does anti differentiating work for questions with ln and e?

I'm reviewing the practice sheet for the final exam. One question is: True or false: antidifferention of 1/2x times dx = ln x^2 + C.

What does the ln mean? Does it just represent 1/2/1, so 2?
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