5.2 definite integrals bernhard reimann. 30 34 0 1
TRANSCRIPT
5.2 Definite Integrals
Bernhard Reimann
30
34
kk=2
50
∑2k
k=2
50
∑3k 2
k=2
100
∑
n2 + 3nk=1
4
∑0
1
When we find the area under a curve by adding rectangles, the answer is called a Rieman sum.
0
1
2
3
1 2 3 4
211
8V t= +
subinterval
partition
The width of a rectangle is called a subinterval.
The entire interval is called the partition.
Subintervals do not all have to be the same size.
→
0
1
2
3
1 2 3 4
211
8V t= +
subinterval
partition
If the partition is denoted by P, then the length of the longest subinterval is called the norm of P and is denoted by .P
As gets smaller, the approximation for the area gets better.
P
( )0
1
Area limn
k kP
k
f c x→
=
= Δ∑ if P is a partition of the interval [ ],a b
→
( )0
1
limn
k kP
k
f c x→
=
Δ∑ is called the definite integral of
over .f [ ],a b
If we use subintervals of equal length, then the length of a
subinterval is:b a
xn
−Δ =
The definite integral is then given by:
( )1
limn
kn
k
f c x→∞
=
Δ∑
→
( )1
limn
kn
k
f c x→∞
=
Δ∑ Leibnitz introduced a simpler notation for the definite integral:
( ) ( )1
limn b
k ank
f c x f x dx→∞
=
Δ =∑ ∫
Note that the very small change in x becomes dx.
→
( )b
af x dx∫
IntegrationSymbol
lower limit of integration
upper limit of integration
integrandvariable of integration
(dummy variable)
It is called a dummy variable because it is integrated out in the final answer. →
( )b
af x dx∫
We have the notation for integration, but we still need to learn how to evaluate the integral.
→
0
1
2
3
1 2 3 4
time
velocity
After 4 seconds, the object has gone 12 feet.
In section 5.1, we considered an object moving at a constant rate of 3 ft/sec.
Since rate . time = distance: 3t d=
If we draw a graph of the velocity, the distance that the object travels is equal to the area under the line.
→
S 4( )= 3 dt 04∫
= 12
211
8v t= +What if:
We could split the area under the curve into a lot of thin trapezoids, and each trapezoid would behave like the large one in the previous example.
It seems reasonable that the distance will equal the area under the curve.
→
0
1
2
3
1 2 3 4x
211
8
dsv t
dt= = +
31
24s t t= +
s =
124
43 +4
26
3s =
The area under the curve2
63
=
We can use anti-derivatives to find the area under a curve!
→
0
1
2
3
1 2 3 4x
Let’s look at it another way:
a x
Let area under the
curve from a to x.
(“a” is a constant)
( )aA x =
x h+
( )aA x
Then:
( ) ( ) ( )a x aA x A x h A x h+ + = +
( ) ( ) ( )x a aA x h A x h A x+ = + −
( )xA x h+
( )aA x h+
→
x x h+
min f max f
The area of a rectangle drawn under the curve would be less than the actual area under the curve.
The area of a rectangle drawn above the curve would be more than the actual area under the curve.
short rectangle area under curve tall rectangle≤ ≤
( ) ( )min max a ah f A x h A x h f⋅ ≤ + − ≤ ⋅
h
( ) ( )min max a aA x h A x
f fh
+ −≤ ≤
→
( ) ( )min max a aA x h A x
f fh
+ −≤ ≤
As h gets smaller, min f and max f get closer together.
( ) ( ) ( )0
lim a a
h
A x h A xf x
h→
+ −= This is the definition
of derivative!
( ) ( )a
dA x f x
dx=
Take the anti-derivative of both sides to find an explicit formula for area.
( ) ( )aA x F x c= +
( ) ( )aA a F a c= +
( )0 F a c= +
( )F a c− =initial value
→
( ) ( )min max a aA x h A x
f fh
+ −≤ ≤
As h gets smaller, min f and max f get closer together.
( ) ( ) ( )0
lim a a
h
A x h A xf x
h→
+ −=
( ) ( )a
dA x f x
dx=
( ) ( )aA x F x c= +
( ) ( )aA a F a c= +
( )0 F a c= +
( )F a c− = A
ax( ) =F x( )−F a( )
Area under curve from a to x = antiderivative at x minus
antiderivative at a. →
( )0
1
limn
k kP
k
f c x→
=
= Δ∑
( )b
af x dx= ∫
=F b( )−F a( )
Area
→
f x( )a
b∫ dx = fnInt f(x),x,a,b( )
On your Ti-84 calculator:
Area from x=0to x=1
0
1
2
3
4
1 2
Example: 2y x=
Find the area under the curve from x=1 to x=2.
2 2
1x dx∫
23
1
1
3x
31 12 1
3 3⋅ − ⋅
8 1
3 3−
7
3=
Area from x=0to x=2
Area under the curve from x=1 to x=2.
→
-1
0
1
Example:
Find the area between the
x-axis and the curve
from to .
cosy x=
0x = 3
2x
π= 2
π
3
2
π
3
2 2
02
cos cos x dx x dxπ π
π−∫ ∫/ 2 3 / 2
0 / 2sin sinx x
π π
π−
3sin sin 0 sin sin
2 2 2
π π π⎛ ⎞ ⎛ ⎞− − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
( ) ( )1 0 1 1− − − − 3=
On the TI-89:
( )( )( )abs cos , ,0,3 / 2x x π∫3=
If you use the absolute value function, you don’t need to find the roots.
+
-
π
-1
0
1
Example:
Find the area between the
x-axis and the curve
from to .
cosy x=
0x = 3
2x
π= 2
π
3
2
π
3
2 2
02
cos cos x dx x dxπ π
π−∫ ∫/ 2 3 / 2
0 / 2sin sinx x
π π
π−
3sin sin 0 sin sin
2 2 2
π π π⎛ ⎞ ⎛ ⎞− − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
( ) ( )1 0 1 1− − − − 3=
On the Ti-84:
3=
If you use the absolute value function, you don’t need to find the roots.
+
-
fnInt abs cos x( )( ),x,0,3π / 2( )
Example2 2
2Evaluate the integral 4 - x dx
−∫
-2 2
3
2
1
-1
A =12π 2( )2 =2π
Example
Evaluate the integral sin x dxπ
−π∫2
1
-1
-2
-2 2
Example
( ) ( ) ( )b
a
Evaluate the integral sin x dx = 0
Note : f x dx = area above axis - area below axis
π
−π∫
∫2
1
-1
-2
-2 2
A =cos(x) −ππ
=cos π( )−cos −π( ) =0
Discontinuous Functions
Some functions with discontinuities are also integrable. A bounded function that has a finite number of points of discontinuity on an interval [a, b] will still be integrable on the interval if it is continuous everywhere else.
2
1
xFind dx
x−∫
Discontinuous Function
2
1
x dx
x
= 1 1 + 1 2
= 1
−
⋅− ⋅
∫
Try this on your calculator using fnInt…
2
1
-1
-2
-2 2
f x( ) = x
x