integrals
DESCRIPTION
5. INTEGRALS. INTEGRALS. Indefinite Integrals. INDEFINITE INTEGRAL. The notation ∫ f ( x ) dx is traditionally used for an antiderivative of f and is called an indefinite integral. Thus, ∫ f ( x ) dx = F ( x ) means F’ ( x ) = f ( x ). INDEFINITE INTEGRALS. - PowerPoint PPT PresentationTRANSCRIPT
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INTEGRALSINTEGRALS
5
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Indefinite Integrals
INTEGRALS
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The notation
∫ f(x) dx is traditionally used for an
antiderivative of f and is called an indefinite
integral.
Thus, ∫ f(x) dx = F(x) means F’(x) = f(x)
INDEFINITE INTEGRAL
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INDEFINITE INTEGRALS
For example, we can write
Thus, we can regard an indefinite integral as representing an entire family of functions (one antiderivative for each value of the constant C).
3 32 2because
3 3
x d xx dx C C x
dx
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INDEFINITE INTEGRALS
Any formula can be verified by differentiating
the function on the right side and obtaining
the integrand.
For instance, 2
2
sec tan
because
(tan ) sec
x dx x C
dx C x
dx
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TABLE OF INDEFINITE INTEGRALS
1
2 2
( ) ( ) [ ( ) ( )]
( ) ( )
( 1)1
sin cos cos sin
sec tan csc cot
sec tan sec csc cot csc
nn
cf x dx c f x dx f x g x dx
f x dx g x dx
xk dx kx C x dx C n
n
x dx x C x dx x C
x dx x C x dx x C
x x dx x C x x dx x C
Table 1
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INDEFINITE INTEGRALS
Thus, we write
with the understanding that it is valid on
the interval (0, ∞) or on the interval (-∞, 0).
2
1 1dx C
xx
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INDEFINITE INTEGRALS
This is true despite the fact that the general
antiderivative of the function f(x) = 1/x2,
x ≠ 0, is:
1
2
1if 0
( )1
if 0
C xxF x
C xx
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INDEFINITE INTEGRALS
Find the general indefinite integral
∫ (10x4 – 2 sec2x) dx
Using our convention and Table 1, we have:
∫(10x4 – 2 sec2x) dx = 10 ∫ x4 dx – 2 ∫ sec2x dx = 10(x5/5) – 2 tan x + C = 2x5 – 2 tan x + C
You should check this answer by differentiating it.
Example 1
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INDEFINITE INTEGRALS
Evaluate
This indefinite integral isn’t immediately apparent in Table 1.
So, we use trigonometric identities to rewrite the function before integrating:
Example 2
2
cos 1 cos
sin sinsin
csc cot csc
d d
d C
2
cos
sind
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The Substitution Rule
In this section, we will learn:
To substitute a new variable in place of an existing
expression in a function, making integration easier.
INTEGRALS
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• Our antidifferentiation formulas don’t tell
us how to evaluate integrals such as
22 1x x dx
Equation 1INTRODUCTION
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• To find this integral, we use the problem-
solving strategy of introducing something
extra.
The ‘something extra’ is a new variable.
We change from the variable x to a new variable u.
INTRODUCTION
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• Suppose we let u = 1 + x2 be the quantity
under
the root sign in the integral below,
Then, the differential of u is du = 2x dx.
INTRODUCTION
22 1x x dx
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• Notice that, if the dx in the notation for
an integral were to be interpreted as
a differential, then the differential 2x dx
would occur in
INTRODUCTION
22 1x x dx
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• So, formally, without justifying our
calculation, we could write:
2 2
3/ 223
2 3/ 223
2 1 1 2
( 1)
x x dx x x dx
udu
u C
x C
Equation 2INTRODUCTION
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• However, now we can check that we have
the correct answer by using the Chain Rule
to differentiate
2 3 2 2 1 232 23 3 2
2
( 1) ( 1) 2
2 1
dx C x x
dx
x x
INTRODUCTION
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• In general, this method works whenever
we have an integral that we can write in
the form
∫ f(g(x))g’(x) dx
INTRODUCTION
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• Observe that, if F’ = f, then
• ∫ F’(g(x))g’(x) dx = F(g(x)) + C
• because, by the Chain Rule,
Equation 3INTRODUCTION
( ( )) '( ( )) '( )dF g x F g x g x
dx
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• If we make the ‘change of variable’ or
‘substitution’ u = g(x),
we have:
'( ( )) '( ) ( ( ))
( )
'( )
F g x g x dx F g x C
F u C
F u du
INTRODUCTION
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• Writing F’ = f, we get:
∫ f(g(x))g’(x) dx = ∫ f(u) du
Thus, we have proved the following rule.
INTRODUCTION
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SUBSTITUTION RULE
• If u = g(x) is a differentiable function
whose range is an interval I and f is
continuous
on I, then
∫ f(g(x))g’(x) dx = ∫ f(u) du
Equation 4
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SUBSTITUTION RULE
• Notice that the Substitution Rule for
integration was proved using the Chain Rule
for differentiation.
• Notice also that, if u = g(x), then du = g’(x)
dx.
So, a way to remember the Substitution Rule is to think of dx and du in Equation 4 as differentials.
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SUBSTITUTION RULE
• Thus, the Substitution Rule says:
• It is permissible to operate with
dx and du after integral signs as if
they were differentials.
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SUBSTITUTION RULE
• Find ∫ x3 cos(x4 + 2) dx
We make the substitution u = x4 + 2.
This is because its differential is du = 4x3 dx, which, apart from the constant factor 4, occurs in the integral.
Example 1
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SUBSTITUTION RULE
• Thus, using x3 dx = du/4 and the Substitution
Rule, we have:
Notice that, at the final stage, we had to return to the original variable x.
3 4 1 14 4
14
414
cos( 2) cos cos
sin
sin( 2)
x x dx u du udu
u C
x C
Example 1
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SUBSTITUTION RULE
• The idea behind the Substitution Rule is to
replace a relatively complicated integral by
a simpler integral.
This is accomplished by changing from the original variable x to a new variable u that is a function of x.
Thus, in Example 1, we replaced the integral ∫ x3cos(x4 + 2) dx by the simpler integral ¼ ∫ cos u du.
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SUBSTITUTION RULE
• The main challenge in using the rule is
to think of an appropriate substitution.
You should try to choose u to be some function in the integrand whose differential also occurs—except for a constant factor.
This was the case in Example 1.
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SUBSTITUTION RULE
• If that is not possible, try choosing u to be
some complicated part of the integrand—
perhaps the inner function in a composite
function.
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• Finding the right substitution is
a bit of an art.
It’s not unusual to guess wrong.
If your first guess doesn’t work, try another substitution.
SUBSTITUTION RULE
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SUBSTITUTION RULE
• Evaluate
Let u = 2x + 1.
Then, du = 2 dx.
So, dx = du/2.
2 1x dxE. g. 2—Solution 1
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Thus, the rule gives:
1 212
3 2
3 213
3 213
2 12
1
2 3/ 2
(2 1)
dux dx u
u du
uC
u C
x C
SUBSTITUTION RULE E. g. 2—Solution 1
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SUBSTITUTION RULE
• Find
Let u = 1 – 4x2. Then, du = -8x dx. So, x dx = -1/8 du and
21 4
xdx
x
1 21 18 82
21 18 4
1
1 4
(2 ) 1 4
xdx du u du
ux
u C x C
Example 3
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SUBSTITUTION RULE
• Calculate ∫ cos 5x dx
If we let u = 5x, then du = 5 dx. So, dx = 1/5 du. Therefore,
15
15
15
cos5 cos
sin
sin 5
x dx u du
u C
x C
Example 4