Integrating exponentials

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amyuser
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#1
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Please can someone explain how to integrate these two expressions and explain how you are supposed to know when you need to use integration by parts and when you don't need it?
1) xe^(-x^2)
2) xe^(4x)
Last edited by amyuser; 3 months ago
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tonyiptony
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For me, there is sort of a "hierarchy" of what integration technique to use.

If you can straight up do the integral (using integral table that you may have learned by heart), do that;
If not, I would think of substitution; (remember substitution exists, it's way more powerful than you think)
If not, then IBP, trig sub, partial fraction, etc., which are the heavy artillery for high school calculus.

For instance, for (1), the first consideration would be whether you can make a substitution, so to turn the integral into an easy one?
It turns out to be yes! Substituting u=-x^2 works.
For (2) however, we can't find a substitution to do just that, then I'll think of using IBP.

For now, just do more and "spot the patterns". To use IBP, usually you can find something that you want to differentiate, and whatever is left is easy to integrate.
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davros
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#3
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#3
(Original post by amyuser)
Please can someone explain how to integrate these two expressions and explain how you are supposed to know when you need to use integration by parts and when you don't need it?
1) xe^(-x^2)
2) xe^(4x)
As the previous poster says, a lot of this is down to practice and experience.

Integration by parts requires a product of two functions. Sometimes this "product" may be non-obvious, for example to integrate arctan x you can treat it as a product where the "second function", dv/dx, is just 1.

For your examples it might look like integration by parts could be applied to both. However, can we actually do this in example 1? We can't integrate e^(-x^2) - in fact it's a famous example of a function that can't be integrated. Therefore we would have to differentiate it. If we did that it would bring down an extra factor of 2x each time. Also we would be integrating x to start with which becomes x^2/2. So our e^(-x^2) is staying the same each time, but the powers of x in front of it are getting bigger. So IBP is making the problem worse, not better! Therefore we look for a substitution. We know how to integrate e^{ku) with respect to u, so something like u = x^2 or u = -x^2 is worth trying. If the "other factor" - in this case x - is a multiple of du/dx then substitution works out nicely

For your second example, we have a product of 2 functions, and in this case we know how to differentiate and integrate both of them., so it looks like IBP is the way to go.
Is there a better choice for which one we take as the "first function" and which one as the "second function"? Well, we can differentiate x to get 1 and integrate e^(4x) to get (1/4)e^(4x), OR we could integrate x to get x^2/2 and differentiate e^(4x) to get 4e^(4x). Hopefully you can see which choice makes life simpler for us and which choice is making things more complicated
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WAGUNGA
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#4
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you use integratin by parts if the expression is a product. In this case, both expressions 1 and 2 are products of x and an exponential function, so you will need to use integration by parts in the steps and probabbly also integration by substitution.
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davros
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(Original post by WAGUNGA)
you use integratin by parts if the expression is a product. In this case, both expressions 1 and 2 are products of x and an exponential function, so you will need to use integration by parts in the steps and probabbly also integration by substitution.
How exactly do you think integration by parts will help with example 1) ?
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WAGUNGA
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#6
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if the fnction is a product, you let one part be u and the other one be dv. For example, in the first task, you may let x be u and the exponential part be dv.
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DFranklin
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(Original post by WAGUNGA)
if the fnction is a product, you let one part be u and the other one be dv. For example, in the first task, you may let x be u and the exponential part be dv.
In which case you're going to have to integrate e^(-x^2). Which is a problem, since there's no way of solving that integral.
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