Integration question, for e Watch

jackhpward
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Need help remembering something from my notes. Will upload pics and circle the bit I'm struggling with. Eg Why does it become 1/(-0.25).. or 1/(-5).. instead of just the denominator?
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jackhpward
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(Original post by jackhpward)
Need help remembering something from my notes. Will upload pics and circle the bit I'm struggling with. Eg Why does it become 1/(-0.25).. or 1/(-5).. instead of just the denominator?
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jackhpward
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RDKGames
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(Original post by jackhpward)
Need help remembering something from my notes. Will upload pics and circle the bit I'm struggling with. Eg Why does it become 1/(-0.25).. or 1/(-5).. instead of just the denominator?
Use a substitution, then it's obvious and in the exam you shoudn't need to think about this.

Let u = -5x and do it.
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Muttley79
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(Original post by jackhpward)
Need help remembering something from my notes. Will upload pics and circle the bit I'm struggling with. Eg Why does it become 1/(-0.25).. or 1/(-5).. instead of just the denominator?
Are you querying where the negative sign comes from or what?

You are using the knowledge of differentiation to integrate.

When we differentiate e^(-5x) we get -5e^(-5x) using the chain rule.

So integrating -5e^(-5x) gives e^(-5x) so to integrate e^(-5x) we need to ....
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jackhpward
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(Original post by Muttley79)
Are you querying where the negative sign comes from or what?

You are using the knowledge of differentiation to integrate.

When we differentiate e^(-5x) we get -5e^(-5x) using the chain rule.

So integrating -5e^(-5x) gives e^(-5x) so to integrate e^(-5x) we need to ....
No I'm questioning why this has become a fraction (easier to see in the second pic). Using the chain rule it should become -0.25 x e^(-0.25)?

I know this is probably a really simple question, but I just can't remember why.
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RDKGames
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(Original post by jackhpward)
Using the chain rule it should become -0.25 x e^(-0.25)?
Don't confuse your differentiation with integration.

You have here found that \dfrac{d}{dx}(e^{-0.25x}) = -0.25e^{-0.25x}

So, integrate both sides and you get \displaystyle e^{-0.25x} = -0.25 \int e^{-0.25x} .dx

Thus \displaystyle \dfrac{1}{-0.25}e^{-0.25x} = \int e^{-0.25x} .dx as required.
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Muttley79
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(Original post by jackhpward)
No I'm questioning why this has become a fraction (easier to see in the second pic). Using the chain rule it should become -0.25 x e^(-0.25)?

I know this is probably a really simple question, but I just can't remember why.
Yes it does when we differentiate but we are integrating. See my example in the previous post.
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jackhpward
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(Original post by RDKGames)
Don't confuse your differentiation with integration.

You have here found that \dfrac{d}{dx}(e^{-0.25x}) = -0.25e^{-0.25x}

So, integrate both sides and you get \displaystyle e^{-0.25x} = -0.25 \int e^{-0.25x} .dx

Thus \displaystyle \dfrac{1}{-0.25}e^{-0.25x} = \int e^{-0.25x} .dx as required.
Thank you! I knew it was something simple that I was missing.
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