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The product rule for derivatives leads to a technique of integration that breaks a complicated integral into simpler parts. Integration by Parts Formula: udv = ...
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[ f ( g ( x ))]' = f '( g ( x )) g '( x ) derivative antiderivative
∫^ f ( g ( x )) g '( x ) dx € F '= f. ∫^ f ( g ( x )) g '( x ) dx^ =^ F ( g ( x ))^ +^ C. composition of functions derivative of Inside function F is an antiderivative of f integrand is the result of differentiating a composition of functions
Integration by Substitution
€ g ( x ) ∫^ f^ ( g ( x )) g '( x ) dx^ =^ ∫ f^ ( u ) du € u = g ( x ) € €^ du^ =^ g '( x ) dx g '( x ) € u = g ( x )
Integration by Substitution
€ ∫^ f^ ( u ) du^ =^ F ( u )^ +^ C € F ( u ) + C = F ( g ( x )) + C
The product rule for derivatives leads to a technique of integration that breaks a complicated integral into simpler parts. Integration by Parts Formula:
hopefully this is a simpler Integral to evaluate given integral that we cannot solve
d dx [ u ( x ) ⋅ v ( x )] = du dx ⋅ v ( x ) + u ( x ) ⋅ dv dx € u ( x ) ⋅ dv dx : u ( x ) ⋅ dv dx
d dx [ u ( x ) ⋅ v ( x )] − du dx ⋅ v ( x )
∫^ u ( x )^ dv dx dx = (^) ∫ d dx [ u ( x ) ⋅ v ( x )] dx − (^) ∫ v ( x ) du dx dx ⇒ (^) ∫ u ( x ) dv = u ( x ) ⋅ v ( x ) − (^) ∫ v ( x ) du
Template : Choose: Compute:
u = € part which gets simpler dv = (^) easy to integrate part after differentiation du = v^ =
Strategy for Integration Method Applies when… Basic antiderivative …the integrand is recognized as the reversal of a differentiation formula, such as Guess-and-check …the integrand differs from a basic antiderivative in that “x” is replaced by “ax+b”, for example Substitution …both a function and its derivative (up to a constant) appear in the integrand, such as Integration by parts …the integrand is the product of a power of x and one of sin x, cos x, and ex, such as …the integrand contains a single function whose derivative we know, such as
Strategy for Integration
impossible to integrate e − x 2 dx 0 1 ∫
easy to integrate impossible to integrate e − x 2 dx 0 1 ∫ ≈^ (^1 −^ x 2 ) dx 0 1 ∫ for x-values near 0
We can obtain a better approximation by using a higher degree Taylor polynomial to represent the integrand. Example: -2.4 -2 -1.6 -1.2 -0.8 -0.4 0 0.4 0.8 1.2 1.6 2 2.
1
e − x 2 dx 0 1 ∫ ≈ ( 1 − x 2
1 2 x 4 − 1 6 x 6 ) dx 0 1 ∫ ≈ x − 1 3 x 3
1 10 x 5 − 1 42 x 7 ⎡ ⎣ ⎢ 0 1 ≈ 0. 74286