Integration of composite functions using substitution, including integrals of the form f(ax+b), and handling definite integrals by changing the bounds or substituting back.
If F(x)=∫f(x)dx, then
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
and let u=g(x), giving us
an integral we can solve more easily:
Then, we substitute g(x) back in to get our desired result of kf(g(x))+C.
When we make a substitution in a definite integral in the form
we need to remember that the bounds are from x=a to x=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
but the integral of a derivative simply returns the original function, so we have
Replacing u′ and v′ with dxdu and dxdv respectively and rearranging terms gives us the result
which we use to perform integration by parts.
Nice work completing Techniques of Integration, here's a quick recap of what we covered:
Exercises checked off
Integration of composite functions using substitution, including integrals of the form f(ax+b), and handling definite integrals by changing the bounds or substituting back.
If F(x)=∫f(x)dx, then
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
and let u=g(x), giving us
an integral we can solve more easily:
Then, we substitute g(x) back in to get our desired result of kf(g(x))+C.
When we make a substitution in a definite integral in the form
we need to remember that the bounds are from x=a to x=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
but the integral of a derivative simply returns the original function, so we have
Replacing u′ and v′ with dxdu and dxdv respectively and rearranging terms gives us the result
which we use to perform integration by parts.
Nice work completing Techniques of Integration, here's a quick recap of what we covered:
Exercises checked off