Refer to the attachment: For the part that is underlined in red, on the right hand side of the equation why does the integral's limits go from -5 to 100 + 100 to 12? It isnt going in sequence (due to the 100 to 12) and that doesnt make sense

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- May 12th 2010, 04:46 AMTsAmESequence of limits of definite integrals
Refer to the attachment: For the part that is underlined in red, on the right hand side of the equation why does the integral's limits go from -5 to 100 + 100 to 12? It isnt going in sequence (due to the 100 to 12) and that doesnt make sense

- May 12th 2010, 10:52 AMskeeter
- May 12th 2010, 01:26 PMTsAmE
How do you know to use the 100 as your c, as you are given an integral with 10 as its limit? How would you know to choose between the 10 and 100?

- May 12th 2010, 01:35 PMskeeter
- May 12th 2010, 01:36 PMAllanCuz
Because it is given in the question.

We are told

$\displaystyle \int_{100}^{-5} f(x) dx = 4 $

Which is the same thing as saying

$\displaystyle - \int_{-5}^{100} f(x) dx = 4 $

Going back to the question we want

$\displaystyle \int_{-5}^{12} f(x) dx $

Well...we know the value from -5 to 100, so lets use that. But to make this equal, we then need to go from 100 to 12! Therefore, we get

$\displaystyle \int_{-5}^{12} f(x) dx = \int_{-5}^{100}f(x)dx + \int_{100}^{12} f(x) dx $

We already know $\displaystyle \int_{-5}^{100} f(x)dx = -4 $ but we don't quite know the integral of 100 to 12...so let's break that guy up in the same way.

$\displaystyle \int_{100}^{12} f(x) dx = \int_{100}^{-10} f(x)dx + \int_{100}^{12} f(x) dx $

Of course by reversing the limits we can find a numerical value.

To re-state...we choose our limits given the definitions in the original question