integrals in physics. biblical reference an area 25,000 cubits long and 10,000 cubits wide will...
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Biblical Reference
An area 25,000 cubits long and 10,000 cubits wide will belong to the Levites, who serve in the temple, as their possession for towns to live in.
Ezekiel 45:5
Why do We Need Integrals?
• When things change in Physics, they create a curved graph.
– Finding the area under a curved graph is difficult using traditional methods.
• An integral calculates the area under a curve.
Vel
ocity
(m
/s)
Time (s)
v(t)
x(t)= ∫ v(t) dt
Simple Example
With a linear function, it is easy to find the area under the curve.
vtx
mss
mx 12)4)(3(
How do you find the distance traveled in 4 seconds for an object moving 3 m/s?
t1= 1s t1= 5s
v =3 m/s
v (m/s)
t(s)
Area = 12 m
Life is not Linear…• The distance
traveled during the time interval between t1 and t2 equals the shaded area under the curve
• How do you calculate this area?
t1
v(t)
t2
v(t +Dt)
v (m/s)
t(s)
Graphs that Curve• You can approximate the
area with a series of rectangles of equal width (time interval) and adding up their areas.– There is a lot of error with
this method, because there are gaps between the rectangles and the curve.
Vel
ocity
(m
/s)
Time (sec)
ttvtxAreat
*)(lim)(0
∆t ∆t∆t∆t ∆t
The Answer… an Integral• When a continuous function is summed, a
new sign is used. It is called an Integral, and the symbol looks like this:
• When you are dealing with a situation where you have to integrate realize:– You are given: the derivative– You want: the original function
• You are working backwards – finding the antiderivative.
How do you take an Integral?• Since an integral is the opposite of a
derivative the steps are:1. Raise the power.
2. Divide by the new power.
3. Add a constant.
• Why the constant?– Remember… the derivative of a constant is
nothing. So, the integral of nothing is a constant.
Cn
xdxx
nn
)1(
)1(
Sample Problem• Start at 2m at t = 0 and start moving with the
following velocity function:
• Find the position function.
526)( 2 tttv
dttttx 526)( 2
CtttCttt
tx
52)10(
5
)11(
2
)12(
6)( 23
)10()11()12(
C2 252)( 23 ttttx
Sample Problem – with LimitsAn object is moving at velocity with respect to time according to the equation v(t) = 2t. What is:
a) The displacement function?
b) The distance traveled from t = 2 s to t = 7 s?
a)
b)
dttdtvtx )2()(
2)( ttx
44927
)2()(
22
272
7
2
7
2tdttdtvtx t
t
t
t
t
t
LIMITS
45 m
8 Simple Derivative Rules
Cn
xdxx
nn
)1()3
)1(
Cxdxx
xy
65
5
3
48
8
Cxxx
dxxx
)8
2
3
3(5
)83(5
23
2
dxxfkdxxkf )()()4
8 Simple Derivative Rules
• Why are there only 8 (i.e. Where are the product and quotient rules)?
– Product and Quotient integration require (often) very complex substitutions. It is very rare in physics that you will be taking an integral of a product or a quotient.