Integration by parts Watch

debbie394
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#1
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Will the second part of integration by parts always be integrated.
By this I mean: uv- integrate v du

So I would only integrate v du
Why is this?
Why not integrate the first half uv?
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heathert99
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it’s a rule dude 🤷🏻*♀️ just follow it
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RedGiant
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Well you still have to integrate dv/dx to get v. You may sometimes also have to use integration by parts within integration by parts, where you have to double integrate.
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0xFFFFail
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So the integration by parts rule was made by realising that when integrating multiple functions that are multiplied by each other, the answer is equivalent to using said formula.
The formula is much easier than painstakingly integrating otherwise (not sure how you'd do that tbh) so that's why we use it.
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brainmaster
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(Original post by heathert99)
it’s a rule dude 🤷🏻*♀️ just follow it
this is my math becomes hard.....a wise person always ask the question "why" and if he isn't answered he won't understand it and can't do it..
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simon0
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This is a rough sketch of the resoning.

The prodcut rule for differentiation is given as:

 \dfrac{d[uv]}{dx} = v \dfrac{du}{dx} + u \dfrac{dv}{dx}.

Therefore, integrating (indefiniely) both sides with respect to x:

 \displaystyle uv = \int v \frac{du}{dx} \, dx + \int u \frac{dv}{dx} \, dx,

with some rearranging:

 \displaystyle \therefore \, \int u \frac{dv}{dx} \, dx = uv - \int v \frac{du}{dx} \, dx .
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randombiochemist
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The integration by parts formula is another integral form of the product rule for differentiation:
For a function y = uv where u and v are each a function of x, dy/dx = u∙dv/dx + v∙du/dx

Integrating both sides thus gives:
∫(dy/dx)dx = ∫u(dv/dx)dx + ∫v(du/dx)dx,
i.e. u(dv/dx)dx = uv - u(dv/dx)dx
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brainmaster
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this is how integration by parts work and derived;
you have what we call the product rule while differentiating uv and we say that;
d (uv)/dx= u*dv/dx + v*du/dx
now we can rearrange this and write it as;
u*dv/dx = d (uv)/dx - v*du/dx

we can now integrate both sides of thus equation and we end up with;

untegral (u*dv/dx) = integral (d (uv)/dx) - integral (v*du/dx).

now if you take any function lets say f (x) Then
if integral (d (f (x))/dx) = f (x)
this is true and u can confirm it by maybe substituting f (x) with x^2
d (x^2)/dx = 2x
integral (d (x^2)/dx) = x^2
so this helps us come up with a statement;
inetgral (d (f (x))/dx) = f (x)

so this is the same.thing that happens with the integral (d (uv)/dx)....it becomes uv ALWAYS that's why we use it instead of going through all the trouble of integrating it and ending up with the same answer.....

so our integration by parts formula;

integral (u*dv/dx) = uv - integral (v*du/dx)

hope you now understand the integration by parts let me know if I helped
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