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What is the antiderivative of #e^(8x)#?

Answer 1

#1/8 e^(8x) +C#

We can go through the steps of integrating by substitution, but some find the following more clear:

We know that #d/dx(e^(8x)) = 8e^(8x)#

That is #8# time more that we want the derivative to be. So, we'll multiply by #1/8# (divide by #8#). #d/dx(1/8e^(8x)) = 1/8*8e^(8x) = e^(8x)#

The general antiderivative is, therefore, #1/8e^(8x)+C#.

Here is the substitution solution

#inte^(8x) dx#

Let #u=8x#, so we get #du = 8 dx# and #dx = 1/8 du#

The integral becomes:

#int e^u * 1/8 du = 1/8 int e^u du = 1/8 e^u +C#

Reversing the substitution gives

#inte^(8x) dx = 1/8e^(8x)+C#

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Answer 2

Aurora Alvarado

The antiderivative of ( e^{8x} ) is ( \frac{1}{8} e^{8x} + C ), where ( C ) is the constant of integration.

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Answer from HIX Tutor

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.