Example We can also evaluate a definite integral by interpretation of definite integral. Ex. Find by interpretation of definite integral. Sol. By the interpretation.
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Example We can also evaluate a definite integral by interpretation
of definite integral. Ex. Find by interpretation of definite integral. Sol. By the interpretation of definite integral, we know the
definite integral is the area of the region under the curve from 0 to a. From the graph, we see the region is a quarter disk with radius a and centered origin. Therefore,
2 2
0
aa x dx
2 2 2
0
1.
4
aa x dx a
Example Ex. By interpretation of definite integral, find
Sol. (1)
(2)
52
3
13
(1) sin (2) 2 . xdx xdx
5
3
3
sin 0
xdx
2
12 3 xdx
Properties of definite integral Theorem(linearity of integral) Suppose f and g are
integrable on [a,b] and are constants, then
is integrable on [a,b] and
Theorem(product integrability) Suppose f and g are integrable on [a,b], then is integrable on [a,b].
, f g
[ ( ) ( )] ( ) ( ) . b b b
a a af x g x dx f x dx g x dx
f g
Properties of definite integral Theorem(additivity with respect to intervals)
Remark In the above property, c can be any number, not necessarily between a and b.
When the upper limit is less than the lower limit in the definite integral, it is understood as
Especially, ( ) 0.a
af x dx
( ) ( ) . a b
b af x dx f x dx
( ) ( ) ( ) . b c b
a a cf x dx f x dx f x dx
Comparison properties of integral 1. If for then
2. If for then
3. If for then
4.
( ) 0f x , a x b ( ) 0.b
af x dx
( ) ( )f x g x , a x b ( ) ( ) . b b
a af x dx g x dx
, a x b( ) m f x M
( ) ( ) ( ). b
am b a f x dx M b a
( ) | ( ) | . b b
a af x dx f x dx
Estimation of definite integral Ex. Use the comparison properties to estimate the definite
integral
Sol. Denote Then when
Letting we get the only critical number
By the closed interval method, we find the range for f(x):
1 2 31
2
1 . x x dx
2 3( ) 1 . f x x x1
[ ,1],2
x2
2 3
2 3( ) .
2 1
x xf x
x x
( ) 0, f x2
.3
x
69 69 1( ) 1, and thus ( ) .
9 18 2
b
af x f x dx
Mean value theorems for integrals Second mean value theorem for integrals Let
g is integrable and on [a,b]. Then
there exists a number such that
Proof. Let Since
we have and
Hence
or By intermediate value theorem
[ , ],f C a b
( ) 0 (or ( ) 0) g x g x[ , ] a b
( ) ( ) ( ) ( ) . b b
a af x g x dx f g x dx
[ , ][ , ]max ( ), min ( ).
x a bx a bM f x m f x ( ) 0,g x( ) 0
b
ag x dx ( ) ( ) ( ) ( ). mg x f x g x Mg x
( ) ( ) ( ) ( ) , b b b
a a am g x dx f x g x dx M g x dx
( ) ( ).
( )
b
ab
a
f x g x dxm M
g x dx
Mean value theorems for integrals First mean value theorem for integrals Let
then there exists a number such that
Remark. We call the mean value of f on [a,b].
[ , ],f C a b
[ , ] a b( ) ( )( ).
b
af x dx f b a
1( ) ( )
b
af f x dx
b a
Example Ex. Suppose and
Prove that such that
Proof. By the first mean value theorem for integrals, there
exists such that Thus
By Rolle’s theorem, such that
[0,1] and (0,1), f C f D1
2
3
3 ( ) (0). f x dx f ( ) 0. f (0,1)
2[ ,1]3
1
2
3
1( ) ( ).
3 f x dx f
1
2
3
(0) 3 ( ) ( ). f f x dx f
(0,1) ( ) 0. f
Function defined by definite integrals with varying limit
Suppose f is integrable on [a,b]. For any given
the definite integral is a number. Letting x vary
between a and b, the definite integral defines a function:
Ex. Find a formula for the definite integral with varying limit
Sol. By interpretation of definite integral, we have
( )x
af t dt
[ , ],x a b
( ) ( ) .x
ag x f t dt
( ) .x
ag x tdt
2 21( ) ( )( ) .
2 2
x
a
x ag x tdt a x x a
Properties of definite integral with varying limit
Theorem(continuity) If f is integrable on [a,b], then the
definite integral with varying limit
is continuous on [a,b].
( ) ( )x
ag x f t dt
The fundamental theorem of calculus (I)
The Fundamental Theorem of Calculus, Part 1 If f is
continuous on [a,b], then the definite integral with varying
limit is differentiable on [a,b] and
Proof
is between x and as and
Therefore,
( ) ( )x
ag x f t dt
( ) ( ) ( ) ( )
x x x x x
a a xg f t dt f t dt f t dt f x
( ) ( ) ( ). x
a
dg x f t dt f x
dx
, x x 0, , x x ( ) ( ).f f x
0 0( ) lim lim ( ) ( ).
x x
gg x f f x
x
Definite integral with varying limits
The definite integral with varying lower limit is
Since we have
The most general form for a definite integral with varying
limits is To investigate its properties,
we can write it into the sum of two definite integrals with
varying upper limit
( ) ( ) ,b
xh x f t dt
( )
( )( ) ( ) .
b x
a xx f t dt
( ) ( ) ( ). b
x
dh x f t dt f x
dx
( ) ( ) , x
bh x f t dt
( ) ( ) ( )
( )( ) ( ) ( ) ( ) .
b x b x a x
a x a ax f t dt f t dt f t dt
Definite integral with varying limits
By the chain rule, we have the formula
( )( ) ( ( )) ( )
b x
af t dt f b x b x
( ) ( ) ( )
( )( ) ( ) ( )
( ( )) ( ) ( ( )) ( )
b x b x a x
a x a af t dt f t dt f t dt
f b x b x f a x a x
Example Ex. Find derivatives of the following functions
Sol.
(2) Let by chain rule,
2
0(1) sin ,
xt tdt 2
1 2(3) ln(1 ) ,x t dt2
2
sin(4) .
x t
xe dt2
0(2) cos ,
xt dt
2 2
0(1) sin sin .
xdt tdt x x
dx
2 2
0 0
1cos cos cos .
2
x ud d dut dt t dt x
dx du dx x
,u x
2
2
1 2 2 4
1(3) ln(1 ) ln(1 ) 2 ln(1 ).
x
x
d dt dt t dt x x
dx dx2 2
2 2 2 4 2sin sin
sin(4) 2 cos .
x x xt t t x x
x a a
d d de dt e dt e dt xe xe
dx dx dx
Example Ex. Find derivative Sol.
Ex. Find if Sol.
22
2(1) ,
x t
xt e dt 2
0(2) (5 ) .
xx t dt
0 0cos 1.
y xte dt tdtdy
dx
222 5 2 2
2(1) 2 8 .
x t x x
x
dt e dt x e x e
dx
2 2 2 2
0 0 0(2) (5 ) (5 ) (5 ) (5 ) .
x x xd dx t dt x t dt t dt x x
dx dx
0 0
coscos 0 cos 0 .
y xt yy
d dy dy xe dt tdt e x
dx dx dx e
Example Ex. Find the limit
Sol. By L’Hospital’s Rule and equivalent substitution,
Question:
2
030
arctanlim .
ln(1 ) x
x
tdt
x
2 2
0 03 3 20 0 0
arctan arctan 2 arctan 2lim lim lim .
ln(1 ) 3 3
x x
x x x
tdt tdt x x
x x x
2
030
arctanlim ?
ln(1 )
x
x
tdt
x
Example Ex. Find the limit
Sol.
2
2
2
0
0 2
0
(1) lim .
x t
xx t
e dt
e dt
2 2 2
2 2
0 0
20 0
2 2 0(1) lim lim 0.
1
x xt x t
x xx x
e dt e e dt
e e
2 3
2
0
0
0
(2) lim .( sin )
x
xx
t dt
t t t dt
3 2
0 0
2 6(2) lim lim 12.
( sin ) 1 cosx x
x x x
x x x x
Example Ex. Suppose b>0, f continuous and increasing on [0,b].
Prove the inequality
Sol. Let
Then F(0)=0 and when
This implies F(t) is increasing, thus
0 02 ( ) ( ) .
b bxf x dx b f x dx
0 0( ) 2 ( ) ( ) .
t tF t xf x dx t f x dx
0 0( ) 2 ( ) ( ) ( ) ( ) ( ) .
t tF t tf t f x dx tf t tf t f x dx
[0, ],t b
( ) ( ) [ ( ) ( )] 0. tf t tf t f t f
( ) (0).F b F
Example Ex. Suppose f is continuous and positive on [a,b]. Let
Prove that there is a unique solution in (a,b) to F(x)=0. Sol.
1( ) ( ) .
( )
x x
a bF x f t dt dt
f t
( ) 0, ( ) 0. ( ) 0 F a F b F
1( ) ( ) 0.
( ) F x f x
f x
Fundamental theorem of calculus (II) The Fundamental Theorem of Calculus, Part 2 If f is
continuous on [a,b] and F is any antiderivative of f, then
Proof Let then g is an antiderivative of f.
So F(x)=g(x)+C. Therefore,
Remark The formula is called Newton-Leibnitz formula
and often written in the form
( ) ( ) ( ). b
af t dt F b F a
( ) ( ) ,x
ag x f t dt
( ) ( ) ( ( ) ) ( ( ) ) ( ) .b a b
a a aF b F a f t dt C f t dt C f t dt
( ) ( ) ( ) . b bb
a aaf t dt F x F x
Example Ex. Evaluate
Sol.
Ex. Find the area under the parabola from 0 to 1. Sol.
24
0tan . xdx
2 24 4 400 0
tan (sec 1) tan 1 .4
xdx x dx x x
2y x13
1 2
00
1.
3 3
xA x dx
Example Ex. Evaluate
Sol.
Ex. Evaluate Sol.
2
0|1 | . x dx
2 1 2
0 0 1|1 | |1 | |1 | x dx x dx x dx
1 22 21 2
0 10 1
(1 ) ( 1) 1.2 2
x xx dx x dx x x
2
0| sin cos | . x x dx
240
4
sin cos cos sin 2( 2 1). x x x x
Example Anything wrong in the following calculation?
33
211
1 1 4.
3
dxx x
Differentiation and integration are inverse
The fundamental theorem of calculus is summarized into
The first formula says, when differentiation sign meets integral sign, they cancel out.
The second formula says, first differentiate F, and then integrate the result, we arrive back to F.
( ) ( ).x
a
df t dt f x
dx
( ) ( ) ( ). b
aF x dx F b F a
Homework 12 Section 5.1: 21
Section 5.2: 22, 37, 53, 59, 67
Section 5.3: 18, 50, 54, 62
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