# How do you evaluate the integral #int 1/(x+1)^(1/3)dx# from -2 to 0?

0

the interval is placed symmetrically about the 2 sided singularity so the areas seem to net off

First, take note of the singularity at x = -1. Out of caution, I'll proceed as follows. Sorry, OTT.

by rule of power

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To evaluate the integral ( \int \frac{1}{(x+1)^{1/3}} , dx ) from -2 to 0, you can use the substitution method. Let ( u = x + 1 ), then ( du = dx ).

When ( x = -2 ), ( u = (-2) + 1 = -1 ).

When ( x = 0 ), ( u = 0 + 1 = 1 ).

Now, the integral becomes ( \int \frac{1}{u^{1/3}} , du ) from -1 to 1.

Integrating ( \frac{1}{u^{1/3}} ) gives ( 3u^{2/3} + C ).

Now, evaluate ( 3u^{2/3} ) at the limits of integration:

At ( u = 1 ): ( 3(1)^{2/3} = 3 ).

At ( u = -1 ): ( 3(-1)^{2/3} = 3 ).

So, the integral ( \int_{-2}^{0} \frac{1}{(x+1)^{1/3}} , dx ) is evaluated as ( 3 - (-3) = 6 ).

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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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