How do you evaluate the definite integral #int v^(1/3)dv# from [-3,3]?
See below.
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To evaluate the definite integral ∫ v^(1/3) dv from -3 to 3, you need to find the antiderivative of v^(1/3) with respect to v first. Then, you evaluate this antiderivative at the upper and lower limits of integration and take the difference.
The antiderivative of v^(1/3) with respect to v is (3/4)v^(4/3).
Now, substitute the upper limit of integration, 3, into the antiderivative and subtract the result of substituting the lower limit of integration, -3:
[(3/4)(3)^(4/3)] - [(3/4)(-3)^(4/3)]
Now, calculate each term:
= [(3/4)(3)^(4/3)] - [(3/4)(3)]
= [(3/4)(9)] - [(3/4)(-3)]
= (27/4) - (-9/4)
= (27/4) + (9/4)
= 36/4
= 9
So, the value of the definite integral ∫ v^(1/3) dv from -3 to 3 is 9.
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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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