# How do you evaluate the definite integral #int -8/3*2^xdx# from -2 to 9?

So, we see that:

Evaluating, this becomes:

Without knowing the formula:

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To evaluate the definite integral ( \int_{-2}^{9} -\frac{8}{3} \cdot 2^x , dx ), we can use the properties of exponential functions and the definite integral.

First, we'll find the antiderivative of ( -\frac{8}{3} \cdot 2^x ). The antiderivative of ( 2^x ) is ( \frac{1}{\ln(2)} \cdot 2^x ). So, the antiderivative of ( -\frac{8}{3} \cdot 2^x ) is ( -\frac{8}{3\ln(2)} \cdot 2^x ).

Now, we'll evaluate the definite integral using the antiderivative:

[ \int_{-2}^{9} -\frac{8}{3} \cdot 2^x , dx = \left[ -\frac{8}{3\ln(2)} \cdot 2^x \right]_{-2}^{9} ]

Plugging in the upper and lower limits:

[ = -\frac{8}{3\ln(2)} \cdot 2^9 - \left( -\frac{8}{3\ln(2)} \cdot 2^{-2} \right) ]

[ = -\frac{8}{3\ln(2)} \cdot 512 + \frac{8}{3\ln(2)} \cdot \frac{1}{4} ]

[ = -\frac{8}{3\ln(2)} \cdot 512 + \frac{2}{3\ln(2)} ]

[ = -\frac{8 \cdot 512}{3\ln(2)} + \frac{2}{3\ln(2)} ]

[ = -\frac{4096}{\ln(2)} + \frac{2}{3\ln(2)} ]

[ = -\frac{4096 - 2}{\ln(2)} ]

[ = -\frac{4094}{\ln(2)} ]

So, ( \int_{-2}^{9} -\frac{8}{3} \cdot 2^x , dx = -\frac{4094}{\ln(2)} ).

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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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