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May 2nd, 2017, 05:56 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, 06:20 PM  #2 
Senior Member Joined: Aug 2012 Posts: 1,971 Thanks: 550 
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 06:40 PM. 
May 2nd, 2017, 06:33 PM  #3 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,842 Thanks: 1068 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, 06:57 PM  #4 
Member Joined: Feb 2017 From: henderson Posts: 36 Thanks: 0  
May 2nd, 2017, 07:07 PM  #5 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,842 Thanks: 1068 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 09:16 PM. 

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