Before: Given f (x ), find f 0 (x ).
Now: Given f 0 (x ), how do you find f (x )?
Example
What function did we differentiate to get: g(x ) = x 5 + 3x − 2?
Solution:
Look at the first term of x 5 .
We got x 5 by differentiating a power function, and by the power rule
(x n )0 = nx n−1 , it looks like we must have differentiated x 6 .
However, if we had differentiated x 6 we would have 6x 5 , and we don’t have a
6 in front of our first term, so the 6 needs to cancel out after we’ve
differentiated.
Thus, it seems we would have to differentiate 61 x 6 to get x 5 .
Likewise for 3x , it seems we must have differentiated 23 x 2 .
Third term is constant and we know if we differentiate x we get 1.
So, we must have differentiated −2x to get −2.
AMAT 217 (University of Calgary)
Fall 2013
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Example
What function did we differentiate to get: g(x ) = x 5 + 3x − 2?
Solution: Putting all of this together gives the following function,
f (x ) =
x6
3x 2
+
− 2x .
6
2
It seems this is the answer, however, we know that the derivative of a constant is
zero and so any of the following will also give g(x ) upon differentiating:
f (x ) =
x 6 3x 2
+
− 2x + 7
6
2
f (x ) =
x 6 3x 2
π
+
− 2x +
6
2
2
f (x ) =
f (x ) =
x 6 3x 2
+
− 2x − 24
6
2
x 6 3x 2
+
− 2x + 1000000000000000000
6
2
In fact, any function of the form,
f (x ) =
x6
3x 2
+
− 2x + C
6
2
will give x 5 + 3x − 2 upon differentiating, where C is any constant.
AMAT 217 (University of Calgary)
Fall 2013
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Terminology
Given f (x ), an anti-derivative of f (x ) is any function F (x ) such that:
F 0 (x ) = f (x ).
If F (x ) is any anti-derivative of f (x ) then the most general anti-derivative of
f (x ) is called an indefinite integral, denoted by,
Z
f (x ) dx = F (x ) + C , C is any constant.
Z
I
The symbol
I
f (x ) is called the integrand
x is called the integration variable
C is called the constant of integration.
I
I
is called the integral symbol
The process of finding the integral is called integration.
AMAT 217 (University of Calgary)
Fall 2013
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Example
Evaluate the following indefinite integral:
Z
x 5 + 3x − 2 dx .
Solution: We just did this problem!!
Since this is asking for the most general anti-derivative we have:
Z
x6
3x 2
+
− 2x + C
x 5 + 3x − 2 dx =
6
2
where C is a constant.
AMAT 217 (University of Calgary)
Fall 2013
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Common mistakes
Don’t drop the dx at the end! Think of the integral and dx as a set of
parenthesis.
Remember to +C at the end of indefinite integrals!
Just like with derivatives, the following will NOT work:
R
Z
Z
Z
Z
f (x ) dx
f (x )
R
f (x )·g(x ) dx 6= f (x ) dx g(x ) dx
and
dx 6=
.
g(x )
g(x ) dx
With derivatives, we had product and quotient rules. For integrals, we have
no such rules.
Integrals are HARD to evaluate because some functions do not have
anti-derivatives!
Example
Z
sin x
dx . That is, there is NO function whose
x
derivative is sin x /x (unless we invent a “new” function).
It is impossible to compute
AMAT 217 (University of Calgary)
Fall 2013
5 / 18
Integration Rules
Integral Rules
Z
Constant Rule:
k dx = kx + C .
Z
Constant Multiple Rule:
Z
kf (x ) dx = k
Z
Sum/Difference Rule:
Z
Power Rules:
x n dx =
k is constant.
Z
f (x ) ± g(x ) dx =
x n+1
+ C , (n 6= −1),
n+1
Z
Exponent Rule:
Z
Trig Rules:
f (x ) dx ,
Z
f (x ) dx ±
Z
1
dx = ln |x | + C , (x 6= 0).
x
e x dx = e x + C .
Z
sin x dx = − cos x +C .
g(x ) dx .
Z
cos x dx = sin x +C .
sec2 x dx = tan x +C .
Proof: By taking derivatives each of the above can be verified.
AMAT 217 (University of Calgary)
Fall 2013
6 / 18
Example
Z
Compute
(x π + 7) dx
Z
(x π + 7)
Z
=
=
AMAT 217 (University of Calgary)
x π dx +
Z
7 dx
x π+1
+ 7x + C
π+1
Fall 2013
7 / 18
Example
If
f 0 (x ) = x 4 + 2x − 8 sin x
then what is f (x )?
Solution: The answer is:
Z
f (x ) = f 0 (x ) dx
Z
=
Z
=
=
4
Z
x dx +
Z
=
x 4 + 2x − 8 sin x dx
x 4 dx + 2
Z
2x dx −
Z
8 sin x dx
Z
x dx − 8
sin x dx
x5
+ x 2 + 8 cos x + C ,
5
where C is a constant.
AMAT 217 (University of Calgary)
Fall 2013
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Example
Z
Evaluate
(z +
√
3
z)(4 − z 2 ) dz
Solution: Remember that we don’t have a rule to deal with integrals of products.
In this case we expand products:
Z
Z
√
2
3
(z + z)(4 − z ) dz =
4z − z 3 + 4z 1/3 − z 7/3 dz
AMAT 217 (University of Calgary)
z2
2
=
4
=
2z 2 −
−
z4
+4
4
z 4/3
4/3
−
z 10/3
+C
10/3
z4
3z 10/3
+ 3z 4/3 −
+C
4
10
Fall 2013
9 / 18
Example
Z
Evaluate
4x 8 − 2x 4 + x 3
dx
x2
Solution:
Remember that we don’t have a rule to deal with integrals of quotients.
Split up fractions:
Z
Z
4x 8 − 2x 4 + x 3
dx
=
4x 6 − 2x 2 + x dx
x2
=
AMAT 217 (University of Calgary)
4x 7
2x 3
x2
−
+
+C
7
3
2
Fall 2013
10 / 18
Differential Equations
Definition - Differential Equation
A differential equation is a mathematical equation for an unknown function of
one (or several) variables that relates the function to its derivatives.
Definition - Order
The order of a differential equation is the largest derivative that appears.
Example
The following equations are examples of differential equations:
y 00 + y 0 = 7 has order 2.
dy
= −y has order 1.
dx
d 5y
dy
4 5 + cos x
= 0 has order 5.
dx
dx
AMAT 217 (University of Calgary)
Fall 2013
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Definition - Solution
A solution to a differential equation is a function that satisfies the equation.
Definition - Initial Condition(s)
Initial condition(s) are a set of points that the solution must satisfy.
Example
For a differential equation involving a function y (x ), initial conditions look like:
y (x0 ) = y0 ,
y 0 (x0 ) = y1 ,
y 00 (x0 ) = y2 ,
. . . etc.
Definition - Initial Value Problem
An initial value problem (IVP) is a differential equation along with a set of
initial conditions.
Example
The following is an IVP:
AMAT 217 (University of Calgary)
d 2y
= sin x ,
dx 2
y 0 (2π) = 0,
y (2π) = 3π.
Fall 2013
12 / 18
Importance of DE’s
Application of Derivatives
If y = f (x ) is a function, then the derivative f 0 (x ) is the rate of change of the
function f (x ).
In a real world problem, we often want to predict future behaviour based on
how current values are related and how they change with respect to each
other (perhaps over time).
Thus, it should make sense that the equations that model behaviour will have
derivatives (rates of change).
Many engineering principles can be described by differential equations.
Differential equations are mathematical tools to model engineering systems:
I
I
I
I
I
hydraulic flow,
heat transfer,
level controller of a tank,
vibration isolation,
electrical circuits, etc.
AMAT 217 (University of Calgary)
Fall 2013
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Example
Sketch various solutions to the differential equation
Solution:
We have:
Z
y =
2x dx = x 2 + C ,
dy
= 2x .
dx
where C is a constant.
This is a parabola translated vertically.
Above we sketch sample solutions for C = 2, C = 0, C = −2 and C = −4.
AMAT 217 (University of Calgary)
Fall 2013
14 / 18
Example
Solve the initial value problem:
dy
= 2x ,
dx
y (0) = 2.
Solution:
Z
As on the last slide, we have:
y =
2x dx = x 2 + C .
Now we use the initial condition y (0) = 2 to solve for C .
This condition says that when x = 0 we have y = 2.
Plugging this into our equation y = x 2 + C gives:
2 = 02 + C
→
C =2
Therefore, the solution to the IVP is y = x 2 + 2.
AMAT 217 (University of Calgary)
Fall 2013
15 / 18
Example
Solve the following initial value problem: y 0 = x 3 ,
y (0) = 1.
Convention: Since x ’s appear in the question, it is assumed that y is a
dy
function of x so that y 0 represents dy
dx (and not dt or something else).
dy
Thus, we have:
= x 3.
dx
This is asking which function has a derivative of x 3 ?
We integrate to get the antiderivative (i.e., the solution y ):
Z
x4
y =
x 3 dx =
+C
4
Now we use the initial condition y (0) = 1 (which is of the form y (x0 ) = y0 ).
This condition says that when x = 0 we have y = 1, that is, the function
needs to go through the point (x0 , y0 ) = (0, 1).
Plugging this into our solution gives:
1=
04
+C
4
Therefore the solution to the IVP is:
AMAT 217 (University of Calgary)
→
y=
C =1
x4
+ 1.
4
Fall 2013
16 / 18
Example
Solve the following initial value problem:
dy
= sec2 t,
dt
y (0) = 1.
Solution:
This is asking which function (of t) has a derivative of sec2 t.
We integrate to get the solution y :
Z
y =
sec2 t dt
= tan t + C
Now we use the initial condition y (0) = 1 (which is of the form y (t0 ) = y0 ).
This condition says that when t = 0 we have y = 1, that is, the function
needs to go through the point (t0 , y0 ) = (0, 1).
Plugging this into our solution gives:
1 = tan 0 + C
Therefore the solution to the IVP is:
AMAT 217 (University of Calgary)
→
C =1
y = tan t + 1.
Fall 2013
17 / 18
Example
Solve the following initial value problem:
d 2y
= sin x ,
dx 2
y 0 (2π) = 0,
y (2π) = 3π.
This is asking which function has a second derivative of sin x and goes through the point
(2π, 3π) such that its derivative goes through the point (2π, 0).
We integrate to get y 0 (since we are given y 00 = sin x ):
y0
Z
=
sin x dx
− cos x + C
=
Now we use the initial condition y 0 (2π) = 0 to solve for C .
This condition says that when x = 2π we have y 0 = 0.
Plugging this into our equation y 0 = − cos x + C gives:
→
0 = −(1) + C
0 = − cos(2π) + C
→
C =1
Therefore, y 0 = 1 − cos x . Now integrating again gives:
Z
y
=
(1 − cos x ) dx
=
x − sin x + D
Now we use the initial condition y (2π) = 3π to solve for D.
This condition says that when x = 2π we have y = 3π.
Plugging this into our equation y = x − sin x + D gives:
→
3π = 2π − 0 + D
3π = 2π − sin(2π) + D
→
D=π
Therefore, the solution to the IVP is y = x − sin x + π.
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Fall 2013
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