# How do you evaluate the definite integral #int (x-x^3)dx# from [0,1]?

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The integral has value

We have by basic integration:

Now find the value using the 2nd fundamental theorem of calculus.

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To evaluate the definite integral ∫(x - x^3) dx from 0 to 1, you integrate the given function with respect to x and then evaluate it at the upper and lower limits of integration, which are 0 and 1, respectively.

∫(x - x^3) dx = [x^2/2 - x^4/4] evaluated from 0 to 1

Plugging in the upper limit (1): = (1^2/2 - 1^4/4)

Plugging in the lower limit (0):

- (0^2/2 - 0^4/4)

Subtract the result of the lower limit from the upper limit: = (1/2 - 1/4) - (0 - 0) = 1/2 - 1/4 = 1/4

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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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