# How do you integrate #int 5^x-3^xdx# from #[0,1]#?

The answer is

Taking log on both sides

Therefore,

Similarly,

Taking log on both sides

Therefore,

Therefore,

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The integral has value

Separating the integrals, we get:

Hopefully this helps!

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To integrate ( \int_{0}^{1} 5^x - 3^x , dx ), you can use the properties of exponents to simplify the integral.

First, split the integral into two separate integrals:

[ \int_{0}^{1} 5^x , dx - \int_{0}^{1} 3^x , dx ]

Now, integrate each term separately:

For ( \int_{0}^{1} 5^x , dx ): [ \int_{0}^{1} 5^x , dx = \left[ \frac{5^x}{\ln(5)} \right]_{0}^{1} = \frac{5^1}{\ln(5)} - \frac{5^0}{\ln(5)} = \frac{5}{\ln(5)} - \frac{1}{\ln(5)} = \frac{4}{\ln(5)} ]

For ( \int_{0}^{1} 3^x , dx ): [ \int_{0}^{1} 3^x , dx = \left[ \frac{3^x}{\ln(3)} \right]_{0}^{1} = \frac{3^1}{\ln(3)} - \frac{3^0}{\ln(3)} = \frac{3}{\ln(3)} - \frac{1}{\ln(3)} = \frac{2}{\ln(3)} ]

Substitute these results back into the original expression:

[ \int_{0}^{1} 5^x - 3^x , dx = \frac{4}{\ln(5)} - \frac{2}{\ln(3)} ]

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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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