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Words 900

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(9.6 in MFE book, 5.5 in e-book)

April 16, 2015

Integration by Substitution(9.6 in MFE book, 5.5 in e-book)

Warm-Up

Let’s ﬁnd

2

2xe x dx.

What we CAN’T do:

2

2xe x dx =

Also note

2x dx ·

ex

2

dx ( (f · g ) =

2

f ·

g)

2

e x dx is NOT equal to 1 e x + C . In fact,

2

d x2

2

[e ] = 2xe x . dx We’ve (inadvertently) discovered

2

2

2xe x dx = e x + C .

Integration by Substitution(9.6 in MFE book, 5.5 in e-book)

Example 1

Suppose we didn’t want to guess-and-check

2

2xe x dx.

Let’s make the substitution u = x 2 . The diﬀerential is du = 2x dx.

Making these subs in the integral gives us:

2

2xe x dx =

2

e x · 2x dx =

e u du

2

= eu + C = ex + C .

Integration by Substitution(9.6 in MFE book, 5.5 in e-book)

Substitution Rule

Theorem

If u = g (x), then f g (x)) · g (x)dx =

f (u) du.

(**) Technically we’re assuming the range of g is an interval I on which f is continuous. In practice, this doesn’t come up often.

Integration by Substitution(9.6 in MFE book, 5.5 in e-book)

Example 2

Find x3 2 + x 4 dx.

Solution: We make the substitution u = 2 + x 4 because du = 4x 3 , which (basically!) appears under the integral sign. u = 2 + x 4 , so du = 4x 3 dx, and

√

√

√ du x 3 2 + x 4 dx =

2 + x 4 · x 3 dx = u 4

1

· 2 u 3/2 + C = 6 (2 + x 4 )3/2 + C .

3

√ d Above we used the fact that dx ( 2 u 3/2 ) = u.

3

=

1

4

Integration by Substitution(9.6 in MFE book, 5.5 in e-book)

Example 2, a note

Alternatively, to ﬁnd x3 we could have let u =

√

2 + x4 . so

du = √

x

3

2+

x4

dx =

=

=

2 + x 4 dx.

2x 3 dx, 2 + x4

2+

x4

3

· x dx =

and

2+

x4

√

2 + x4 3

·√

· x dx

2 + x4

x3 dx 2 + x4 du u3

(2 + x 4 )3/2 u2 ·

=

+C =

+ C.

2

6

6

(2 + x 4 ) · √

Integration by Substitution(9.6 in MFE book, 5.5 in e-book)

Example 3

Find

2x − 4e −4x dx. x 2 + e −4x

Solution: The numerator is the derivative of the denominator. u = x 2 + e −4x , du = (2x − 4e −4x ) dx,

2x − 4e −4x dx = x 2 + e −4x

2x − 4e −4x dx = x 2 + e −4x

1 du = ln |u| + C u = ln |x 2 + e −4x | + C = ln(x 2 + e −4x ) + C .

(above, |x 2 + e −4x | = x 2 + e −4x since x 2 + e −4x > 0)

Integration by Substitution(9.6 in MFE book, 5.5 in e-book)

Deﬁnite Integration

We can use substitution in deﬁnite integrals, but we have to change the limits of integration: If u = g (x) then b g (b)

f (g (x)) · g (x) dx = a f (u) du. g (a)

(**) Technically we’re assuming g is continuous on [a, b] and f is continuous on the range of g .

Integration by Substitution(9.6 in MFE book, 5.5 in e-book)

Example 4

Find

4

Solution: If u =

√

4

1

e x

√ dx = x √

1

√

e x

√ dx. x x, then du =

√

√ 4 eu

1

· 2 du =

1

√

2 x

2 u 1 2e

dx, so du = 2e u

2

1

= 2e 2 − 2e.

Integration by Substitution(9.6 in MFE book, 5.5 in e-book)

Example 4, picture

√

x

√

On the left is the graph of e x for x ∈ [1, 4].

On the right is the graph of 2e u for u ∈ [1, 2]. They look diﬀerent, but have the same area.

Integration by Substitution(9.6 in MFE book, 5.5 in e-book)

Example 5

A jet accelerates along the runway and takes oﬀ. It consumes fuel t2 at a rate of f (t) = (t+1)3 kL per minute. How much fuel does the jet consume in the ﬁrst 3 minutes?

Solution: By the Evaluation Theorem, the amount consumed from

3

t2 t = 0 to t = 3 minutes is 0 (t+1)3 dt. We make the substitution u = t + 1 to make the denominator nicer, then we expand.

3

0

t2 dt =

(t + 1)3

4

=

1

4

1

(u − 1)2 du = u3 1

2

1

− 2+ 3 u u u 4

1

u 2 − 2u + 1 du u3

du = ln |u| +

2

1

− 2 u 2u

4

1

2 1

1

16 1

48

33

= [ln(4)+ − ]−[ln(1)+2− ] = [ln(4)+ − ]−

= ln(4) − .

4 32

2

32 32 32

32

Thus, the jet consumes ln(4) −

33

32

≈ 0.355 kL.

Integration by Substitution(9.6 in MFE book, 5.5 in e-book)

Example 6

Find

1 dx. x · ln(x) · ln(ln x)

Solution: If we let u = ln(ln x), then du =

1

dx = x ln(x) ln(ln x)

1 ln x

·

1 x dx, so:

1

1

dx = ln(ln x) x ln x

1 du u

= ln |u| + C = ln | ln(ln x)| + C .

Integration by Substitution(9.6 in MFE book, 5.5 in e-book)…...

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