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Techniques of Integration
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The reverse chain rule, integration by substitution, integration by parts, additional strategies
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Integrating f(ax+b)
AHL AI 5.11
If F(x)=∫f(x)dx, then
∫f(ax+b)dx=a1F(ax+b)🚫
Integration by substitution
AHL AI 5.11
Integrating a composition of functions f(g(x)) requires us to divide by g′(x), so it is easier to find the anti-derivative of anything of the form g′(x)f′(g(x)) by first dividing by g′(x).
In symbols, we use the known fact
∫kg′(x)f′(g(x))dx=kf(g(x))+C🚫
and let u=g(x), giving us
∫kg′(x)f′(g(x))dx=k∫f′(u)du,🚫
an integral we can solve more easily:
k∫f′(u)du=kf(u)+C.
Then, we substitute g(x) back in to get our desired result of kf(g(x))+C.
Substitution and Integral Bounds
AHL AI 5.11
When we make a substitution in a definite integral in the form
∫abkg′(x)f′(g(x))dx
we need to remember that the bounds are from x=a to x=b:
∫abkg′(x)f′(g(x))dx =k∫x=ax=bf′(u)du =[kf(u)]x=ax=b
We then have two choices:
Plug x=a and x=b into u to find the bounds in terms of u.
Plug u(x) back in and use the bounds a→b.
Recall the product rule: (uv)′=u′v+uv′. Then, integrating both sides with respect to x yields
∫(uv)′dx=∫u′vdx+∫uv′dx,
but the integral of a derivative simply returns the original function, so we have
uv=∫u′vdx+∫uv′dx.
Replacing u′ and v′ with dxdu and dxdv respectively and rearranging terms gives us the result
∫udxdvdx=uv−∫vdxdudx,
which we use to perform integration by parts.
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