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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 = ...

Typology: Study Guides, Projects, Research

2021/2022

Uploaded on 09/27/2022

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The$Chain$Rule$and$Integration$by$
Substitution$
Recall:$
The$chain$rule$for$derivatives$allows$us$to$
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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

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 ) dxF '= 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 ) duu = 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 )^ +^ CF ( 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 dxv ( x ) + u ( x ) ⋅ dv dxu ( x ) ⋅ dv dx : u ( x ) ⋅ dv dx

d dx [ u ( x ) ⋅ v ( x )] − du dxv ( 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 ex 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 ex 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

ex 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