# What is the area enclosed by #r=2sin(4theta+(11pi)/12) # between #theta in [pi/8,(pi)/4]#?

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To find the area enclosed by the polar curve ( r = 2\sin(4\theta + \frac{11\pi}{12}) ) between ( \theta = \frac{\pi}{8} ) and ( \theta = \frac{\pi}{4} ), you can use the formula for finding the area enclosed by a polar curve:

[ A = \frac{1}{2} \int_{\theta_1}^{\theta_2} r^2 d\theta ]

where ( r ) is the polar function and ( \theta_1 ) and ( \theta_2 ) are the limits of integration.

First, find the expression for ( r^2 ):

[ r^2 = (2\sin(4\theta + \frac{11\pi}{12}))^2 ]

Now, integrate ( r^2 ) with respect to ( \theta ) over the given interval:

[ A = \frac{1}{2} \int_{\frac{\pi}{8}}^{\frac{\pi}{4}} (2\sin(4\theta + \frac{11\pi}{12}))^2 d\theta ]

After integrating, you will have the area enclosed by the given polar curve between ( \frac{\pi}{8} ) and ( \frac{\pi}{4} ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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