# How do you find the indefinite integral of #t^(9/2)+8t^(1/2)-8t^(-1/2)dt#?

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To find the indefinite integral of (t^{9/2} + 8t^{1/2} - 8t^{-1/2}) with respect to (t), you can use the power rule for integration. The power rule states that the integral of (t^n) with respect to (t) is (\frac{t^{n+1}}{n+1}), where (n) is any real number except -1. Applying this rule to each term of the given expression:

[ \begin{aligned} \int t^{9/2} , dt & = \frac{t^{9/2+1}}{9/2+1} = \frac{2}{11}t^{11/2} + C, \ \int 8t^{1/2} , dt & = 8 \cdot \frac{t^{1/2+1}}{1/2+1} = 8 \cdot \frac{2}{3}t^{3/2} + C = \frac{16}{3}t^{3/2} + C, \ \int -8t^{-1/2} , dt & = -8 \cdot \frac{t^{-1/2+1}}{-1/2+1} = -8 \cdot 2t^{1/2} + C = -16t^{1/2} + C. \end{aligned} ]

Therefore, the indefinite integral of (t^{9/2} + 8t^{1/2} - 8t^{-1/2}) with respect to (t) is (\frac{2}{11}t^{11/2} + \frac{16}{3}t^{3/2} - 16t^{1/2} + C), where (C) is the constant of integration.

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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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