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Lecture 12

MATH 2B Lecture Notes - Lecture 12: Product RulePremium


Department
Mathematics
Course Code
MATH 2B
Professor
ERJAEE, G.
Lecture
12

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02/01/2019
MATH 2B - Lecture 12 - Integration by Parts
Every differentiation rule has a corresponding integration rule. The integration by parts
corresponds to the product rule for differentiation.
Product Rule: [𝑓(π‘₯)𝑔(π‘₯)]β€² = 𝑓′(π‘₯)𝑔(π‘₯) + 𝑓(π‘₯)𝑔′(π‘₯)
Integral:
∫ [𝑓(π‘₯)𝑔(π‘₯)]′𝑑π‘₯
= ∫ 𝑓′(π‘₯)𝑔(π‘₯)𝑑π‘₯ + ∫ 𝑓(π‘₯)𝑔′(π‘₯)𝑑π‘₯
or 𝑓(π‘₯)𝑔(π‘₯) + 𝑐 = ∫ 𝑓′(π‘₯)𝑔(π‘₯)𝑑π‘₯ + ∫ 𝑓(π‘₯)𝑔′(π‘₯)𝑑π‘₯
● Here we ignore the constant β€œc” on the left-hand side because the two indefinite
integrals on the right will give us another two constants later
● So we have: 𝑓(π‘₯)𝑔(π‘₯) = ∫ 𝑓′(π‘₯)𝑔(π‘₯)𝑑π‘₯ + ∫ 𝑓(π‘₯)𝑔′(π‘₯)𝑑π‘₯
● We rearrange to obtain:
∫ 𝑓(π‘₯)𝑔′(π‘₯)𝑑π‘₯ = 𝑓(π‘₯)𝑔(π‘₯) βˆ’ ∫ 𝑓′(π‘₯)𝑔(π‘₯)𝑑π‘₯
Now we simplify the formula by using new notations
● Let 𝑒 = 𝑓(π‘₯) and 𝑣 = 𝑔(π‘₯) then 𝑑𝑒 = 𝑓′(π‘₯)𝑑π‘₯ 𝑑𝑣 = 𝑔′(π‘₯)𝑑π‘₯
● The formula becomes
∫ 𝑒 𝑑𝑣 = 𝑒 β‹… 𝑣 βˆ’ ∫ 𝑣 𝑑𝑒
● That means instead of solving ∫ 𝑒 𝑑𝑣 directly, we can use integration by parts to
arrive at the right-hand side, and solve ∫ 𝑣 𝑑𝑒 which could be simpler
Example
∫ π‘₯𝑠𝑖𝑛π‘₯ 𝑑π‘₯
Solution:
𝑒 = π‘₯ 𝑑𝑣 = 𝑠𝑖𝑛π‘₯ 𝑑π‘₯
𝑑𝑒 = 𝑑π‘₯ 𝑣 = = π‘π‘œπ‘ π‘₯
∫ π‘₯𝑠𝑖𝑛π‘₯ 𝑑π‘₯ = π‘₯(βˆ’π‘π‘œπ‘ π‘₯) βˆ’ ∫ βˆ’ π‘π‘œπ‘ π‘₯ 𝑑π‘₯
u dv u v v du
= βˆ’π‘₯π‘π‘œπ‘ π‘₯ + ∫ π‘π‘œπ‘ π‘₯ 𝑑π‘₯
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