# How do you find the area under the graph of #f(x)=sin(x)# on the interval #[-pi,pi]# ?

For this question to have to set up a definite integral.

Be careful with all of those double negatives which resolve to positives.

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To find the area under the graph of ( f(x) = \sin(x) ) on the interval ([- \pi, \pi]), you would integrate ( \sin(x) ) with respect to ( x ) from (-\pi) to (\pi).

[ \text{Area} = \int_{-\pi}^{\pi} \sin(x) , dx ]

Using the fundamental theorem of calculus, the antiderivative of ( \sin(x) ) is ( -\cos(x) ). So, integrating ( \sin(x) ) from (-\pi) to (\pi) yields:

[ \int_{-\pi}^{\pi} \sin(x) , dx = [-\cos(x)]_{-\pi}^{\pi} ]

Evaluating ( -\cos(x) ) at (\pi) and (-\pi) and subtracting them gives:

[ (-\cos(\pi)) - (-\cos(-\pi)) = (-(-1)) - (-(-1)) = 1 - 1 = 2 ]

Thus, the area under the graph of ( f(x) = \sin(x) ) on the interval ([- \pi, \pi]) is (2).

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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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