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May 2nd, 2017, 05:56 PM   #1
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Question 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
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May 2nd, 2017, 06:20 PM   #2
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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.
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May 2nd, 2017, 06:33 PM   #3
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$$\int e^{-\cos\theta}\sin\theta\,d\theta=\int e^u\frac{du}{d\theta}d\theta=\int e^u\,du=e^u+C=e^{-\cos\theta}+C$$
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May 2nd, 2017, 06:57 PM   #4
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$$\int e^{-\cos\theta}\sin\theta\,d\theta=\int e^u\frac{du}{d\theta}d\theta=\int e^u\,du=e^u+C=e^{-\cos\theta}+C$$
how do you use the Fundamental Theorem of Calculus on the problem?
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May 2nd, 2017, 07:07 PM   #5
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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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