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$$r = e^{- \theta/4}$$ $$\pi /2 \leq \theta \leq \pi$$

I know the formula is

$$\int_a^b \frac{1}{2} r^2 d\theta$$

$$\int_{\pi/2}^{\pi} \frac{1}{2} (e^{- \theta/4})^2 d\theta$$

From here I cannot figure out an easy way to integrate this.

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3 Answers 3

up vote 1 down vote accepted

Note that $(e^{-\theta/4})^2=e^{-\theta/2}$. It is a special case of the general exponential rule $(a^b)^c=a^{bc}$.

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I was attempting to work my way through how to do that so I did $2^3^2$ on my calculator and it gave me 512 which means that it is $2^9$ which means that I have to square the exponent. What went wrong? –  user138246 Jul 2 '12 at 19:36
    
@Jordan It may be because of the order of operation of the calculator. $$(2^3)^2=8^2=64=2^{2*3}=2^6$$ –  Argon Jul 2 '12 at 19:53
    
@Jordan Your calculator chose to give you 2^(3^2); this is not the same as (2^3)^2. In general the calculator will have its own precedence rules, and without understanding them I'd be very careful about trusting what it says about composing operations. –  Steven Stadnicki Jul 2 '12 at 19:53

The integral you compute should is:

$$\frac{1}{2} \int_{\pi/2}^{\pi}\left(e^{-\theta/4}\right)^2\, d\theta=\frac{1}{2} \int_{\pi/2}^{\pi}e^{-\theta/2}\, d\theta$$

Using the exponent rule $(a^b)^c=a^{bc}$. Letting $u=-\theta/2$, $\theta=-2u$, $d\theta=-2\, du$

$$\frac{1}{2}\int_{\pi/2}^{\pi}e^{-\theta/2}\, d\theta=-\int_{-\pi/4}^{-\pi/2}e^u\, du=-\left[e^u\right]^{-\pi/2}_{-\pi/4}=e^{-\pi/2}(e^{\pi/4}-1)$$

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Hint: $(x^a)^b=x^{ab}$. After that, it falls right out, either through FTC or a substitution.

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