# What is the area enclosed by #r=theta^2-2sintheta # for #theta in [pi/4,pi]#?

Area

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To find the area enclosed by the curve (r = \theta^2 - 2\sin(\theta)) for (\theta) in the interval (\left[\frac{\pi}{4}, \pi\right]), you can integrate the formula for the area enclosed by a polar curve:

[A = \frac{1}{2} \int_{\alpha}^{\beta} [f(\theta)]^2 , d\theta]

where (f(\theta)) represents the polar function and (\alpha) and (\beta) are the starting and ending angles, respectively.

For this particular curve, (f(\theta) = \theta^2 - 2\sin(\theta)), and the interval is from (\frac{\pi}{4}) to (\pi).

So, the area enclosed by the curve is:

[A = \frac{1}{2} \int_{\frac{\pi}{4}}^{\pi} (\theta^2 - 2\sin(\theta))^2 , d\theta]

You would need to evaluate this integral to find the exact area.

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