The$Chain$Rule$and$Integration$by$
Substitution$
Recall:$
The$chain$rule$for$derivatives$allows$us$to$
differentiate$a$composition$of$functions:$
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$
$
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[f(g(x))]' =f'(g(x))g'( x)
derivative$
antiderivative$
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The Chain Rule and Integration by Substitution, Study Guides, Projects, Research of Differential and Integral Calculus
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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The Chain Rule and Integration by
Substitution
Recall :
The chain rule for derivatives allows us to
differentiate a composition of functions:
[ f ( g ( x ))]' = f '( g ( x )) g '( x ) derivative antiderivative
The Chain Rule and Integration by
Substitution
Suppose we have an integral of the form
where
Then, by reversing the chain rule for derivatives,
we have
∫^ 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
Algorithm :
1. Let where is the part causing
problems and cancels the remaining x
terms in the integrand.
2. Substitute and into the
integral to obtain an equivalent (easier!)
integral all in terms of u.
€ 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
Algorithm :
3. Integrate with respect to u, if possible.
4. Write final answer in terms of x again.
€ ∫^ f^ ( u ) du^ =^ F ( u )^ +^ C € F ( u ) + C = F ( g ( x )) + C
The Product Rule and
Integration by Parts
The product rule for derivatives leads to a technique of integration that breaks a complicated integral into simpler parts. Integration by Parts Formula:
udv = uv − vdu
hopefully this is a simpler Integral to evaluate given integral that we cannot solve
The Product Rule and
Integration by Parts
Deriving the Formula
Start by writing out the Product Rule:
Solve for
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 )
The Product Rule and
Integration by Parts
Deriving the Formula
Simplify:
∫^ 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
Integration by Parts
Template : Choose: Compute:
udv = uv − vdu
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
What if the integrand does not have a formula
for its antiderivative?
Example:
impossible to integrate e − x 2 dx 0 1 ∫
Integration Using Taylor Polynomials
We approximate the function with an
appropriate Taylor polynomial and then
integrate this Taylor polynomial instead!
Example:
easy to integrate impossible to integrate e − x 2 dx 0 1 ∫ ≈^ (^1 −^ x 2 ) dx 0 1 ∫ for x-values near 0
Integration Using Taylor Polynomials
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