Integration by parts/Fact
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Integration by parts
Integration by parts
Let
f
,
g
:
[
a
,
b
]
⟶
R
{\displaystyle f,g\colon [a,b]\longrightarrow \mathbb {R} }
denote
continuously differentiable
functions
.
Then
∫
a
b
f
(
t
)
g
′
(
t
)
d
t
=
f
g
|
a
b
−
∫
a
b
f
′
(
t
)
g
(
t
)
d
t
.
{\displaystyle {}\int _{a}^{b}f(t)g'(t)\,dt=fg|_{a}^{b}-\int _{a}^{b}f'(t)g(t)\,dt\,.}
Proof
,
Write another proof
![{\displaystyle f,g\colon [a,b]\longrightarrow \mathbb {R} }](../_assets_/eb734a37dd21ce173a46342d1cc64c92/62d73c59b59bcbf77e12fcb056dfc8687ee91d08.svg)
