Course: Accelerated Engineering Calculus I
Instructor: Michael Medvinsky
12.2 The Definite Integrals (5.2)
Def: Let f(x) be defined on interval [a,b]. Divide [a,b] into n subintervals of equal
b−a
, so x0 = a, x1 = a + Δx, x j = a + jΔx, xn = b . Let x *j be an arbitrary (sample)
n
points such that x *j ∈ ( x j−1, x j ) . Then the definite integral of f from a to b is
width Δx =
b
∫
a
n
f ( x ) dx = lim ∑ f ( x *j ) Δx provided that the limit exists. If it douse exist, we say that f is
n→∞
j=1
integrable on [a,b].
Notes:
•
∫
is an integral sign, f(x) is an integrand and a, b are lower and upper limits
of the integral respectively. Evaluating\calculating the integral is called
integration.
• The integral is not dependend on x, i.e.
b
∫
b
f ( x ) dx =
a
∫
a
b
f (t ) dt =
∫ f (r ) dr
a
• If f(x)>0 in [a,b], then an integral represent the area that lies under f(x). For
f(x)<0 in [a,b], the integral represent –A of -f(x). If f changes signs, then it
represent the difference between areas of negative and positive regions.
• The Reimann sum can be equivalently defined using intervals of unequal width.
Ex 2.
Evaluate
3
∫ x − 2 dx
.
0
Solution: Sketch the graph of x-2 and figure out that the function is crossing x-axis,
therefore the integral is the difference of areas (of 2 triangles). Thus
3
1
1
1
∫ x − 2 dx = 2 (1⋅1) − 2 (2 ⋅ 2) = 2 − 2 = −1.5 .
0
Thm: If f is continuous on [a,b], or if it has finite number of jump discontinuities,
then f is integrable.
Thm: If f is integrable on [a,b] then
b
∫
a
x0 = a, x1 = a + Δx, x j = a + jΔx, xn = b .
n
f ( x ) dx = lim ∑ f ( x j ) Δx where Δx =
n→∞
j=1
b−a
and
n
Course: Accelerated Engineering Calculus I
12.2.1
Ex 3.
Instructor: Michael Medvinsky
Evaluating Integrals
n
n
n
b
b
∫ 1dx =
∫ dx = lim ∑1⋅ Δx = lim ∑ Δx = lim Δx∑1 = lim nΔx = lim n
a
n→∞
a
n→∞
j=1
n
b
n→∞
j=1
n→∞
j=1
n
n
n→∞
b−a
= (b − a)
n
∫ x dx = lim ∑ x Δx = lim ∑(a + jΔx ) Δx = lim ∑(aΔx + j (Δx ) ) =
Ex 4.
n→∞
a
j
n→∞
j=1
n→∞
j=1
2
j=1
n
n
%
(
% b − a % b − a (2 n ( n +1) (
%
2
2 n ( n +1) (
*=
= lim '' aΔx∑1 + ( Δx ) ∑ j ** = lim ' anΔx + ( Δx )
+'
* = lim ' an
*
n→∞
n→∞
& n )
2 ) n→∞ '&
n
2 *)
&
&
j=1
j=1 )
2
2
2
%
b − a ) ( n +1) (
b − a)
b − a)
% n +1 (
(
(
(
'
*
= lim ' a ( b − a ) +
= a (b − a) +
lim '
=
* = a (b − a) +
n→∞
n
2 *)
2 n→∞ & n )
2
&
=
2ab − 2a 2 + b 2 − 2ab + a 2 b 2 − a 2
=
2
2
n
b
Ex 5.
∫x
2
n
2
(
2
)
dx = lim ∑ x Δx = lim ∑ ( a + jΔx ) Δx = lim ∑ a 2 + 2ajΔx + ( jΔx ) Δx
n→∞
a
n
2
j
n→∞
j=1
n→∞
j=1
j=1
n
n
% n
(
%
2
3
2 n ( n +1)
3 n ( n +1) ( 2n +1) (
= lim '' ∑ a 2 Δx + ∑ 2aj ( Δx ) + ∑ j 2 ( Δx ) ** = lim ' na 2 Δx + 2a ( Δx )
+ ( Δx )
*=
n→∞
2
6
)
& j=1
) n→∞ &
j=1
j=1
3
%
b − a ) ( n +1) ( 2n +1) (
2 ( n +1)
(
2
'
*=
= a ( b − a ) + lim ' a ( b − a )
+
2
*
n→∞
n
6
n
&
)
2
= a (b − a) + a (b − a)
2
(b − a)
+
3
(b − a)
3
6
2
= a 2 (b − a) + a (b − a) +
∫x
a
n
(
6
3
(n +1) (2n +1)
n2
n→∞
3
2n 2 + 3n +1
2
( b − a) = b3 − a3
2
=
a
b
−
a
+
a
b
−
a
+
( ) ( )
n→∞
n2
3
3 3
lim
n
b
Ex 6.
lim
n
n
3
(
2
n→∞
n→∞
j=1
2
n→∞
j=1
3
4
j=1
)
= lim ∑ a 3Δx + 3a 2 j ( Δx ) + 3aj 2 ( Δx ) + j 3 ( Δx ) =
n→∞
j=1
2
2
3
4
&
b − a ) n ( n +1) ( 2n +1) ( b − a ) & n ( n +1) ) )+
(
3
2 ( b − a ) n ( n +1)
(
= lim a ( b − a ) + 3a
+ 3a
+
(
+ +=
2
3
4
n→∞ (
n
2
n
6
n
2
'
* *
'
3
4
2
&
a ( b − a ) ( n +1) ( 2n +1) ( b − a ) & n +1 ) )
3
2 n +1
3
2
= lim (( a ( b − a ) + a ( b − a )
+
+
(
+ ++ =
2
n→∞
'
* *
2
n
2
n
4
n
'
4
3
2
3
(b − a) = b 4 − a 4
= a (b − a) + a 2 (b − a) + a (b − a) +
2
4
4 4
3
3
)
dx = lim ∑ x 3j Δx = lim ∑ ( a + jΔx ) Δx = lim ∑ a 3 + 3a 2 jΔx + 3a ( jΔx ) + ( jΔx ) Δx =
Course: Accelerated Engineering Calculus I
Instructor: Michael Medvinsky
12.2.2
Midpoint Rule
It is sometime usable to estimate integral numerically. Instead of evaluating limits one
choose finite (small) n and x *j =
Ex 7.
x j−1 + x j
, then use the definition.
2
The exact solution, as shown in previous example is
3
2
3
3
24
=4
4
3
1 " 0 +1 / 2 % 1 " 1 / 2 +1 % 1 " 1+ 3 / 2 % 1 " 3 / 2 + 2 %
x 3 dx = $
' + $
' + $
' + $
' =
2# 2 & 2# 2 & 2# 2 & 2# 2 &
∫
0
3
3
3
3
1 " 1 % 1 " 3 % 1 " 5 % 1 " 7 % 1 " 1 + 27 + 125+343 % 31
= $ ' + $ ' + $ ' + $ ' = $
'=
& 8
2#4& 2#4& 2#4& 2#4& 2#
43
Ex 8.
π
Approximate
∫ sin x dx = 2 using Midpoint Rule, compare to “end point”
0
rule (aka x = x j ).
*
j
π
∫ sin x dx ≈ π sin
•
0
π
π
π
=π
2
π
π
π
π
π
π
∫ sin x dx ≈ π sin π = 0
0
∫ sin x dx ≈ 2 sin 4 + 2 sin
0
•
π
3π π 1 π 1
π
=
+
=
≈ 2.22
4 2 2 2 2
2
π
π
∫ sin x dx ≈ 2 sin 2 + 2 sin π = 2 + 0 = 2 ≈ 1.5708
0
12.2.3
1.
Properties of Definite Integral
b
∫
a
2.
n
n
a
b−a
a−b
f ( x ) dx = lim ∑ f ( x j ) Δx = lim ∑ f ( x j )
= − lim ∑ f ( x j )
= − ∫ f ( x ) dx
n→∞
n→∞
n→∞
n
n
b
j=1
j=1
j=1
n
a
∫ f ( x ) dx = lim ∑ f ( x )
n→∞
a
3.
n
j
j=1
a−a
=0
n
n
b
n
n→∞
a
j
n→∞
j=1
j
∫ { f ( x ) ± g ( x )} dx = lim ∑{ f ( x ) ± g ( x )} Δx =
n→∞
a
n
j
n→∞
n
n→∞
j=1
b
b
∫ f ( x ) dx ± ∫ g ( x ) dx
a
j=1
n
b
j
j=1
= lim ∑ f ( x j ) Δx ± lim ∑ g ( x j ) Δx =
5.
a
j=1
n
b
4.
b
∫ c dx = lim ∑ cf ( x ) Δx = c lim ∑ f ( x ) Δx = c ∫ dx = c (b − a)
n
a
b
∫ cf ( x ) dx = lim ∑ cf ( x j ) Δx = c lim ∑ f ( x j ) Δx = c ∫ f ( x ) dx
a
n→∞
j=1
n→∞
j=1
a
Course: Accelerated Engineering Calculus I
6.
b
∫
c
f ( x ) dx =
a
∫
b
f ( x ) dx +
a
c
∫ f ( x ) dx
c
b
b
∫ dx + ∫ dx = (c − a) + (b − c) = b − a =
Ex 9.
a
c
c
b
∫ x dx + ∫
a
Instructor: Michael Medvinsky
c
∫ dx
a
# c2 a2 & # b2 c2 & b2 a2
x dx = % − ( + % − ( = − =
$2 2' $2 2' 2 2
7. If f ( x ) ≥ 0, ∀x ∈ [ a, b] ⇒
b
∫ x dx
a
b
∫ f ( x ) dx ≥ 0
a
f ( x ) ≥ g ( x ), ∀x ∈ [ a, b] ⇒ f ( x ) − g ( x ) ≥ 0
8.
b
⇒
∫
b
f ( x ) − g ( x ) dx =
a
∫
f ( x ) dx −
a
b
Ex 10.
Estimate
b
∫ g ( x ) dx ≥ 0 ⇒
∫
a
9. m ≤ f ( x ) ≤ M ⇒ ∫ m dx ≤
a
b
b
3
a
b
∫ g ( x ) dx
a
b
∫ f ( x ) dx ≤ ∫ M dx ⇒ m (b − a) ≤ ∫ f ( x ) dx ≤ M (b − a)
a
a
1
∫x
b
f ( x ) dx ≥
− x 2 +1dx =
0
a
11
≈ 0.916 : We would like to bound the function,
12
i.e. we first find global min and max value of the integrand. Let f ( x ) = x 3 − x 2 +1 ,
the first derivative test give us f ' ( x ) = 3x 2 − 2x = 0 ⇔ x = 0, 2 , we next compare
3
! 2 $ 23
value of f at these and end points. f ( 0) = 1; f (1) = 1; f #" 3 &% = 27 ≈ 0.852 . Thus
1
0.852 ≤ f ( x ) ≤ 1 ⇒ 1⋅ 0.852 ≤
∫ f ( x ) dx ≤ 1⋅1
0
12.3 Evaluating Definite Integrals (5.3)
Evaluation theorem: Let f be continuous on [a,b] and let F be arbitrary
antiderivative of f , i.e. F’=f, then
b
∫ f ( x ) dx = F (b) − F (a)
a
In order to see why the theorem above works we consider dividing [a,b] into n
subintervals with the following end points a = x0 , x1,..., xn = b and Δx = x j+1 − x j =
next use Mean Value Theorem to see that
F ( x j ) − F ( x j−1 )
x j − x j−1
b−a
. We
n
= f ( x *j ) where x *j ∈ #$ x j−1 , x j %& ,
Course: Accelerated Engineering Calculus I
Instructor: Michael Medvinsky
thus F ( x j ) − F ( x j−1 ) = f ( x *j ) Δx . Therefore:
F ( b) − F ( a ) = F ( xn ) − F ( x0 ) = F ( xn ) − F ( xn−1 ) + F ( xn−1 ) +... + −F ( x1 ) + F ( x1 ) − F ( x0 ) =
n
= f ( xn* ) Δx +... + f ( x1* ) Δx = ∑ f ( x *j ) Δx
j=1
n
Note: F ( b) − F ( a) = lim F ( b) − F ( a) = lim ∑ f ( x *j ) Δx =
n→∞
n→∞
j=1
Notation:
b
∫ f ( x ) dx = F ( x )
a
Ex 11.
b
∫ c dx = cx
a
Ex 12.
Ex 14.
a
a
a
= F ( b) − F ( a )
(
)
= b−a c
b
b
1
1
x dx = x 2 = b2 − a 2
2 a 2
(
∫
a
Ex 13.
b
b
b
∫ f ( x ) dx .
)
1
⎧2
1 n+1
1
1
n +1
n
∫−1 x dx = n + 1 x −1 = n + 1 1 − ( −1) = n + 1 1 + ( −1) = ⎨⎩0
1
(
n
π
∫ sin x dx = −cos x
π
0
)
(
)
n = 2m
n = 2m + 1
= (−cos π ) − (−cos0 ) = 1− cos π = 1− (−1) = 2
0
ln π
∫ (
2
) (
0
Ex 15.
ln π
=
)
∫ e (cos
t
2
ln π
∫
0
ln π
2
t
∫ e (cost − sin t ) + (sin t + cost )
2
dt =
0
) (
)
t − 2cost sin t + sin 2 t + cos 2 t + 2cost sin t + sin 2 t dt =
0
=
2
e t cost − e t sin t + e t sin t + e t cost dt =
2e 2t dt = 2
ln π
t
∫ e dt =
0
2e t
ln π
0
(
12.3.1
Indefinite Integral
Def: The common notation of antiderivative is
∫ f ( x ) dx = F ( x )
)
= 2 π −1
∫ f ( x ) dx
means F ' ( x ) = f ( x )
Note that the difference between indefinite integral ∫ f ( x ) dx and definite integral
b
∫ f ( x ) dx
is that the former is a function whereas the later is a number. The connection
a
is give by Evaluation theorem:
b
∫ f ( x ) dx = ∫ f ( x ) dx
a
b
a
Course: Accelerated Engineering Calculus I
Instructor: Michael Medvinsky
Table of indefinite integrals:
1.
2.
∫ cf ( x ) dx = c ∫ f ( x ) dx
∫ f ( x ) + g ( x ) dx = ∫ f ( x ) dx + ∫ g ( x ) dx
1 n+1
x +C
n +1
3.
∫x
4.
∫ x dx = ln x + C
5.
n
dx =
∫e
9.
∫ cos
1
11. ∫
x
dx = e + C
2
x
dx = tan x + C
1
dx = cot x + C
sin 2 x
1
dx = tan −1 x + C
2
x +1
12. ∫
ax
a
dx
=
+C
6. ∫
ln a
x
7.
∫ sin x dx = −cos x + C
10. ∫
1
x
8.
1
1− x
2
dx = sin −1 x + C
∫ cos x dx = sin x + C
Ex 1.
∫
(2x + 7) dx = 2 ∫ x dx +
Ex 2.
∫
(3x + 5) 2 dx =
Ex 3.
Ex 4.
Ex 5.
Ex 6.
∫
∫
1
7 dx = 2 x 2 + 7x + C
2
(9x 2 + 30x + 25) dx = 3 ∫ 3x dx +15 ∫ 2x dx + 25 ∫ dx =
= 3x 3 +15x 2 + 25x + C
dx
1
= tan x + C
2
2cos x 2
1 − cos ( 2 x )
x sin ( 2 x )
2
∫ sin xdx = ∫ 2 dx = 2 − 4 + C
∫
dx
=
1+ cos 2x
∫
1
∫ sin 2 x cos3xdx = 2 ∫ sin ( − x ) + sin (5x ) dx =
∫ tan
2
t = ∫ 1+
1
= t + tan t + C
cos 2 t
cos x cos (5 x )
−
+C
2
10