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 February 4th, 2016, 12:59 AM #1 Member   Joined: Jan 2016 From: / Posts: 43 Thanks: 1 Fundamental theorem of calculus question. here : I just stumbled upon this khan academy video, explaining fundamental theorem of calculus. I am sure than mr. Khan got the right result, but I'm just curious, Couldn't we just swap the bounds of the given expression, get the negative form and evaluate it with the help of the fundamental theorem of calculus? I tried to evaluate it and got a different result, so my question is why this doesn't work? thanks!
 February 4th, 2016, 02:33 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra Can you write out the expression?
February 4th, 2016, 04:23 AM   #3
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I can try!

here I hope it's attached right!
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 February 4th, 2016, 07:54 AM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra I had to watch the video to try to understand what you mean, and I'm still not sure. Are you suggesting that we write: $$F(x) = \int \limits_x^{x^2} {\cos t \over t} \, \mathrm d t = -\int \limits_{x^2}^x {\cos t \over t} \, \mathrm d t$$ and then try to apply the Fundamental Theorem of Calculus to the final expression? The point about the Fundamental Theorem of Calculus is that says that if $$F(x) = \int \limits_a^x f(t) \, \mathrm d t$$ then $$F'(x) = f(x)$$ for any constant $a$. In this example, we don't have a constant limit - both limits are in the variable $x$, so we have to introduce the constant in order to use the theorem.
 February 4th, 2016, 08:03 AM #5 Member   Joined: Jan 2016 From: / Posts: 43 Thanks: 1 Exactly what I meant! Sure.. the thing breaks at the constant. And you're right x is a variable! It all makes sense. Thanks once again!
February 4th, 2016, 03:25 PM   #6
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Quote:
 Originally Posted by SuperNova1250 I tried to evaluate it and got a different result ...
What did you try?

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