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 May 2nd, 2017, 06:56 PM #1 Member   Joined: Feb 2017 From: henderson Posts: 36 Thanks: 0 Use the Fundamental Theorem of Calculus to calculate the definite integral. Give an e So I am having a problem with using the fundamental theorem for this problem but I might have messed up with the substitution part I know I took the derivative and I know its wrong so I tried the anti way May 2nd, 2017, 07:20 PM #2 Senior Member   Joined: Aug 2012 Posts: 2,424 Thanks: 759 I didn't follow your work but at the bottom you have $e^{\sin x} \cdot \sin x$. That can't be right. If you differentiate $e^{\cos x}$ you'll get $e^{\cos x} \cdot \sin x$ give or take some minus signs I was too lazy to account for. But once you do that you'll have the correct antiderivative. Ok let me be unlazy. $\frac{d}{dx} e^{-\cos x} = e^{-\cos x} \cdot (-1)(-\sin x) = e^{-\cos x} \cdot \sin x$. This is exactly your integrand so the definite integral is $e^{-\cos x} \biggr \rvert_0^{\frac{\pi}{2}} = e^0 - e^1 = 1 -e$ if I did that right. I should mention that I didn't use a formal substitution, I just eyeballed the integrand and saw that it was set up to be easy. Last edited by Maschke; May 2nd, 2017 at 07:40 PM. May 2nd, 2017, 07:33 PM #3 Global Moderator   Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,974 Thanks: 1155 Math Focus: Elementary mathematics and beyond $$\int e^{-\cos\theta}\sin\theta\,d\theta=\int e^u\frac{du}{d\theta}d\theta=\int e^u\,du=e^u+C=e^{-\cos\theta}+C$$ May 2nd, 2017, 07:57 PM   #4
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 Originally Posted by greg1313 $$\int e^{-\cos\theta}\sin\theta\,d\theta=\int e^u\frac{du}{d\theta}d\theta=\int e^u\,du=e^u+C=e^{-\cos\theta}+C$$
how do you use the Fundamental Theorem of Calculus on the problem? May 2nd, 2017, 08:07 PM #5 Global Moderator   Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,974 Thanks: 1155 Math Focus: Elementary mathematics and beyond FTOC: $$\int_a^bf(x)\,dx=F(b)-F(a)$$ where $F(x)$ is the antiderivative of $f(x)$. Can you work from there? Last edited by greg1313; May 2nd, 2017 at 10:16 PM. Tags calculate, calculus, definite, fundamental, give, integral, theorem 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 Johanovegas Calculus 4 January 17th, 2016 06:03 AM Mr Davis 97 Calculus 1 April 16th, 2015 05:06 PM bonildo Calculus 2 September 11th, 2014 11:00 AM Jhenrique Calculus 6 April 12th, 2014 09:34 PM Aurica Calculus 1 June 10th, 2009 06:39 PM

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