# What is the arc length of #f(x)=cosx-sin^2x# on #x in [0,pi]#?

That doesn't simplify easily so computer solution is:

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To find the arc length of a function ( f(x) ) on an interval ([a, b]), the formula is:

[ L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} , dx ]

For ( f(x) = \cos(x) - \sin^2(x) ) on ( x \in [0, \pi] ):

[ \frac{dy}{dx} = -\sin(x) - 2\sin(x)\cos(x) ]

[ L = \int_{0}^{\pi} \sqrt{1 + \left(-\sin(x) - 2\sin(x)\cos(x)\right)^2} , dx ]

[ L \approx 4.525 ]

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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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