MATH1051 Lecture Notes - Lecture 11: Product Rule
Lecture #11 – 7.1 - Integration by Parts
What is the derivative of (x)g(x)?f
●Product rule
●f(x)g(x)) f(x)g(x) f(x)g(x)( ′= ′+ ′
●This kind of answer is difficult to see if you want to take the integral of something
●How can we rewrite it?
●(f(x)g(x)) (x)g(x) (x)g(x)
∫
′= ∫
f′+ ∫
f′
○First take the integral
○(x)g(x) (x)g(x) (x)g(x)f= ∫
f′+ ∫
f′
■The integral of the left side is the same as (x)g(x)f
○(x)g(x) f(x)g(x) (x)g(x)
∫
f′= − ∫
f′
■Rearrange
●What does this mean?
○dv uv du
∫
u= − ∫
v
○Replacing , you can get this formula(x) with u,and g(x) with vf
○This is integration by parts
Example #1
●e dx
∫
xx
○, which means that xu = u 1dxd =
■From the integral, replace with which parts look easiest to deriveu
○, then v ed = x v=ex
■From the integral, replace with which parts look easiest to integratevd
○dv uv du
∫
u= − ∫
v
○e x(e)
∫
xx= x− ∫
ex
○xe e = x− x
○(x) Cex− 1 +
How do we know which to set to and which to set to ?u vd
●LIATE
○Logs
○Inverse Logs
○Algebra
○Trig
○Exponentials
●Going down the list means that it is easier to take the integral of
○In other words, logs are the easiest to take the derivative of while exponentials
are the easiest to take the integral of
Example #2
●sin(2x)dx
∫
x
○, so xu = u 1dxd =
○, so v sin(2x)dxd = in(2x)dxv = ∫
s
■Here, we have to use substitution to further integrate v
■ 2xdx, ( )w dxw = 2
1=
■in(u)du
2
1∫
s
■cos(u) cos(2x)
2
−1 = − 2
1
■ cos(2x)v= − 2
1
Document Summary
Lecture #11 7. 1 - integration by parts. What is the derivative of (x)g(x)? f f(x)g(x)) This kind of answer is difficult to see if you want to take the integral of something. How can we rewrite it? f (x)g(x) f(x)g (x) (f(x)g(x)) First take the integral (x)g(x) f f (x)g (x) f (x)g(x) (x)g(x) f f (x)g (x) The integral of the left side is the same as (x)g(x) f f (x)g (x) f(x)g(x) (x)g(x) f . What does this mean? u dv uv v du f. This is integration by parts (x) with u, and g(x) with v. Example #1 x x e dx u = x. From the integral, replace with u which parts look easiest to derive d = x e v. From the integral, replace with vd which parts look easiest to integrate v du u uv dv x x = e x x e ex 1 + ) c (x x(e ) xe x ex.
