# How do you evaluate the definite integral #int (t^3-9t)dt# from [-1,1]?

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To evaluate the definite integral ( \int_{-1}^{1} (t^3 - 9t) , dt ), you first need to find the antiderivative of the integrand, which is ( \frac{t^4}{4} - \frac{9t^2}{2} ).

Then, you substitute the upper limit (1) and the lower limit (-1) into the antiderivative and subtract the result of substituting the lower limit from the result of substituting the upper limit.

So, ( \int_{-1}^{1} (t^3 - 9t) , dt = \left[ \frac{t^4}{4} - \frac{9t^2}{2} \right]_{-1}^{1} = \left[ \left( \frac{1^4}{4} - \frac{9 \cdot 1^2}{2} \right) - \left( \frac{(-1)^4}{4} - \frac{9 \cdot (-1)^2}{2} \right) \right] ).

Simplify the expression:

( = \left( \frac{1}{4} - \frac{9}{2} \right) - \left( \frac{1}{4} - \frac{9}{2} \right) ).

( = \left( -\frac{17}{2} \right) - \left( -\frac{17}{2} \right) ).

( = -\frac{17}{2} + \frac{17}{2} ).

( = 0 ).

Therefore, ( \int_{-1}^{1} (t^3 - 9t) , dt = 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.

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