Integration

Integration

BY

J. MBEDU

7.1 - Antiderivatives

If F(x) = 10x, then F’(x) = 10. F(x) is the antiderivative of f(x) = 10

If F(x) = x2, F’(x) = 2x. F(x) is the antiderivative of f(x) = 2x

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Antiderivatives

Find the antiderivative of f(x) = 5x4

Work backwards (from finding the derivative)

The antiderivative of f(x)=F’(x) is x5

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Antiderivatives

In the example we just did, we know that F(x) = x2 is not the only function whose derivative is f(x) = 2x

G(x) = x2 + 2 has 2x as the derivative

H(x) = x2 – 7 has 2x as the derivative

For any real number, C, the function F(x)=x2 + C has f(x) as an antiderivative

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There is a whole family of functions having 2x as an antiderivative

This family differs only by a constant

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Since the functions

G(x) = x2 F(x) = x2 + 2 H(x) = x2 – 7 differ only by a constant, the slope of the tangent line remains the same

The family of antiderivatives can be represented by F(x) + C

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The family of all antiderivaties of f is indicated by

f(x)dxIntegral sign Integrand

This is called the indefinite integral and is the most general antiderivative of f

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Example

Using this new notation, the dx means the integral of f(x) with respect to x

If we write a gets treated as a constant and x as the variable

If we write x gets treated as the constant

22ax dx x C

2 22ax da a x C xa C

22ax dx a(2x)dx ax C

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Finding the Antiderivative

Finding the antiderivative is the reverse of finding the derivative. Therefore, the rules for derivatives leads to a rule for antiderivatives

Example:

So

5 4dx 5x

dx

4 55x dx x C 22 FEBRUARY 2014 201246302 J.MBEDU 10

Rules for Antiderivatives

Power Rule:

Ex:

Ex:

1

1

( 1 exp

1

)

n

n

for any real number n

add to the onent and divide by that num

xx dx C

n

ber3

341

3 1 4

t

t dt

Ct

12

2

11

1

t

Ct

dt t dt Ct

You can always check your answers by taking the derivative!

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FOR YOU

1.

2.

u du3

22

3u C

dx x C

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Rules for Finding Antiderivatives

Constant Multiple and Sum/Difference:

( ) ( )

k f x dx k f x dx

for any real number k

( ) ( ) ( ) ( )f x g x dx f x g x dx

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Examples

43

3

4

2 24

2

2

v

vv C

v

vv

d

d C

You do:

5

12dz

z4

3C

z

23 4 5z z dz 3 22 5z z z C

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Example

2

1 1

2 2

1xdx

x x

2 1xx

2 1x

dxx x

3 1

2 2x x dx

Now that we have rewritten the integral, we can find the antiderivative

5 1

2 2

5 1

2 2

x xC

5 1

2 22

25

x x C

First, rewrite the integrand

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Recall

Previous learning:

If f(x) = ex then f’(x) = ex

If f(x) = ax then f’(x) = (ln a)ax

If f(x) = ekx then f’(x) = kekx

If f(x) = akx then f’(x) = k(ln a)akx

This leads to the following formulas:

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Indefinite Integrals of Exponential

Functions

, 0

ln

, 0(ln )

x x

kxkx

xx

kxkx

e dx e C

ee dx C k

ka

a dx Caa

a dx C kk a

This comes from the chart on P. 434

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Examples

9 99 t t te dt e dt e C 9

9

9

tt e

e dt C

55 443 3

5

4

uu e

e du C

55

4443

5

12

5

u ue e CC

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You Do

52 x dx

52

5(ln2)

x

C

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Indefinite Integral of x-1

Note: if x takes on a negative value, then lnx will be undefined. The absolute value sign keeps that from happening.

1 1lnx dx dx x C

x

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Example

4l1

n4

4dx dxx

xx

C

You Do:

25 xe dxx

21

5ln2

xx e C

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Example

looks like the chain rule and product rule.

But using differentials and substitution we’ll find the antiderivative

43 22x 1 6x dx

4 43 2 3 22x 1 6x dx 2x 1 6x dx

u du

4 = u du22 FEBRUARY 2014 201246302 J.MBEDU 22

Example Con’t

Now use the power rule

Substitute (2x3 + 1) back in for u:

4 u du5u

C5

43 2

53

2x 1 6x dx2x 1

C5

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You Do

Find

723x 4 6xdx

726x 3x 4 dx

7u du u du

87 u

u du C8

823x 4C

8

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Choosing u

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du

We haven’t needed the du in the past 2 problems, but that’s not always the case. The du happened to have already appeared in the previous examples.

Remember, du is the derivative of u.

43 22x 1 6x dx

723x 4 6xdx22 FEBRUARY 2014 201246302 J.MBEDU 26

Example

Find

Let u = x3 + 1, then du = 3x2dx

There’s an x2 in the problem but no 3x2, so we need to multiply by 3

Multiplying by 3 changes the problem, so we need to counteract that 3 by also multiplying by 1/3

2 3x x 1dx

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Example

3 21x 1 3x dx

3

2 3 2 31x x 1 dx 3x x 1 dx

3

1

21 1u du u du

3 3

33 322 21 u 1 2 2

C u C u C33 3 3 92

3

3 22x 1 C

9

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Example

Find

u = x2 + 6x, so du = (2x + 6)

22

x 3dx

x 6x

2 22 2

x 3 2 x 31dx dx

2x 6x x 6x

2

1 du

2 u

121 1 u 1

u du C C2 2 1 2u

2

1C

2 x 6x

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THANK YOU

THE END22 FEBRUARY 2014 201246302 J.MBEDU 30

Education

Integration