Reverse Substitution
Mark Stankus
January 4, 2014
Mark Stankus
Reverse Substitution
January 4, 2014
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Table of Contents
1
What is it?
2
Problem 1 (Calculus II immediately)
3
Problem 2 (7.3)
Mark Stankus
Reverse Substitution
January 4, 2014
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What is it?
Table of Contents
1
What is it?
2
Problem 1 (Calculus II immediately)
3
Problem 2 (7.3)
Mark Stankus
Reverse Substitution
January 4, 2014
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What is it?
When performing the substitution u = f (x), if you are able to compute
x in terms of u, then you have two ways to compute dx.
Mark Stankus
Reverse Substitution
January 4, 2014
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What is it?
When performing the substitution u = f (x), if you are able to compute
x in terms of u, then you have two ways to compute dx.
1
Compute du = f 0 (x)dx and then solve for dx
Mark Stankus
Reverse Substitution
January 4, 2014
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What is it?
When performing the substitution u = f (x), if you are able to compute
x in terms of u, then you have two ways to compute dx.
1
Compute du = f 0 (x)dx and then solve for dx
2
Solve for x and then compute dx.
Mark Stankus
Reverse Substitution
January 4, 2014
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Problem 1 (Calculus II immediately)
Table of Contents
1
What is it?
2
Problem 1 (Calculus II immediately)
3
Problem 2 (7.3)
Mark Stankus
Reverse Substitution
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Problem 1 (Calculus II immediately)
Problem 1 (Calculus II immediately)
Z
Mark Stankus
1
√ dx
x x
Reverse Substitution
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Problem 1 (Calculus II immediately)
Problem 1 (Calculus II immediately)
Z
1
√ dx
x x
We could do this by using an algebraic identity.
Mark Stankus
Reverse Substitution
January 4, 2014
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Problem 1 (Calculus II immediately)
Problem 1 (Calculus II immediately)
Z
1
√ dx
x x
We could do this by using an algebraic identity.
Let’s use reverse substitution instead. Let u =
Mark Stankus
Reverse Substitution
√
x.
January 4, 2014
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Problem 1 (Calculus II immediately)
Problem 1 (Calculus II immediately)
Z
1
√ dx
x x
We could do this by using an algebraic identity.
Let’s use reverse substitution instead. Let u =
√
x.
So x = u2 and dx = 2 u du.
Mark Stankus
Reverse Substitution
January 4, 2014
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Problem 1 (Calculus II immediately)
Problem 1 (Calculus II immediately)
Z
1
√ dx
x x
We could do this by using an algebraic identity.
Let’s use reverse substitution instead. Let u =
√
x.
So x = u2 and dx = 2 u du.
Z
1
√ dx =
x x
Mark Stankus
Z
1
2 u du =
u2 u
Z
2
du =
u2
Reverse Substitution
Z
2 u−2 du etc.
January 4, 2014
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Problem 2 (7.3)
Table of Contents
1
What is it?
2
Problem 1 (Calculus II immediately)
3
Problem 2 (7.3)
Mark Stankus
Reverse Substitution
January 4, 2014
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Problem 2 (7.3)
Problem 2 (7.3)
Z √
Mark Stankus
1 + ex
dx
ex
Reverse Substitution
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Problem 2 (7.3)
Problem 2 (7.3)
Z √
1 + ex
dx
ex
Let’s use reverse substitution with u = ex .
Mark Stankus
Reverse Substitution
January 4, 2014
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Problem 2 (7.3)
Problem 2 (7.3)
Z √
1 + ex
dx
ex
Let’s use reverse substitution with u = ex .
u = ex
x = ln(u)
Mark Stankus
dx =
1
du
u
Reverse Substitution
January 4, 2014
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Problem 2 (7.3)
Problem 2 (7.3)
Z √
1 + ex
dx
ex
Let’s use reverse substitution with u = ex .
u = ex
x = ln(u)
Z √
Mark Stankus
dx =
1 + ex
dx =
ex
1
du
u
Z √
1+u 1
du =
u
u
Reverse Substitution
Z √
1+u
etc.
u2
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