How do you find the definite integral for: #cos(x)# for the intervals #[-pi, pi]#?
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To find the definite integral of cos(x) over the interval [-π, π], you can use the properties of trigonometric functions and integration techniques. The integral of cos(x) from -π to π is equal to the integral of cos(x) from 0 to π plus the integral of cos(x) from -π to 0.
Using the property of evenness of cosine function, the integral of cos(x) from -π to 0 is equal to the integral of cos(x) from 0 to π. Therefore, the definite integral of cos(x) over the interval [-π, π] is twice the integral of cos(x) from 0 to π.
Integrating cos(x) from 0 to π yields sin(x), evaluated from 0 to π. So, sin(π) - sin(0).
Since sin(π) is 0 and sin(0) is 0, the definite integral of cos(x) over the interval [-π, π] is 0.
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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.
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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