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INTRODUCTION
Integral Calculus is just as important in Physics, and will be
used throughout the
Introductory Physics course. This is a quick review of some
basic integration concepts.
This manual is not meant to teach you everything about
Integration, but rather to give you
a solid base for learning more on your own.
INTEGRATION Getting Started
Integration is denoted by the sign
Just as with differential calculus, you must be given something
to integrate with respect to
(w.r.t.). You can tell what you are integrating with respect to
by looking at what follows
the d term behind the integral sign.
For example, y = 2x dx means you are integrating 2x w.r.t. x
Or s = 2t dt means you are integrating 2t w.r.t. t
There are 2 types of integrals: indefinite integrals and
definite integrals. We will start by
looking at indefinite integrals.
Indefinite Integrals
Indefinite integrals can be considered as Anti-Derivates i.e.
integration is the inverse of
differentiation. So if 2x is the differential of x2, then x2 is
the anti-differential or the
integral of 2x. i.e. If y = x2 then dy/dx = 2x
therefore: 2x dx = x2
Now, because an infinite number of functions can give you the
same differential, you must
add an arbitrary constant (+ c) to every anti-derivative or
integral.
For example, the following three functions when differentiated
would all give 2x
for the solution:
1) y = x2 dy/dx = 2x
2) y = x2 1 - dy/dx = 2x
3) y = x2 dy/dx = 2x
SO 2x dx = x2 + c
With that said, lets begin to explore the rules of
integration.
Learning Points
Just as with
differential calculus,
you must be given
something to
integrate with respect
to.
You will be able to
spot an indefinite
integral as it has no
upper and lower
limits attached to the
integral sign. (If you dont know what this means, dont worry. By
the end of this manual you will understand).
When you do an
indefinite integral
add an arbitrary
constant to your
answer.
arbitrary constant
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BASIC RULES OF INTEGRATION
a. The Constant Rule
The integral of a constant is the constant times the variable
being integrated with respect
to. So: n dx = nx + c
Practice the concept: 5 df = 5f + c
dy = 1 dy = y + c
4t dx = 4tx + c
b. The Power Rule
To integrate a variable raised to a power (except if the power
is -1), you add one to the
power and divide by the new power.
So, for example, xn dx = )1(
)1(
n
x n + c
Lets practice this concept.
c. Multiplied and Divided Constants
If a variable is multiplied or divided by a constant, the
multiplied or divided constant stays
with the integral. In fact, you can move the multiplied constant
outside of the integral.
Example: 5x2 dx = 5 x2 dx = 3
5 3x + c
Or: 33
1
x dx =
3
1x-3 dx =
6
2x + c
Now you try the examples on your right.
Remember that all
other variables, than
that being integrated
with respect to, are
treated as constants.
(Check out the last
example on the left).
TIP:
It may help to rewrite
some functions before
integrating.
E.g. 1/p2 dp = p-2 dp = -1/p
Sample problems
1. 2 dp
2. dt
3. x3 dx
4. 4y dy
5. 8t-3 dt
Answers:
1. 2p + c
2. t + c
3. x4/4 + c
4. 2y2 + c
5. -4t-2 + c
Question 1
x dx = x1 dx
= )11(
)11(x + c
= 2
2x+ c
Question 2
s-3 ds = )13(
)13(s + c
= 22
1
s + c
Question 3
F dF = )1
2
1(
)12
1(
F + c
= 3
2 23
F + c
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d. Polynomials
To integrate polynomials, integrate each portion of the
polynomial with respect to the
specified variable. In other words, the integral of a sum (or
difference) is the sum (or
difference) of its integrals.
Example: 3x3 + 4x2 - x- dx = 3x3 dx + 4x2 dx - x- dx
= 3 x3 dx + 4 x2 dx - x- dx
= 34
4x + 4
3
3x -
2/1
2
1
x
= 4
3 4x +
3
4 3x - 2x + c
Note: Simplification
It is always best to simplify an equation before attempting to
solve it. By doing this, you
could save yourself a lot of trouble
Example: dxx
xx
)5(
562 = dx
x
xx
)5(
)1)(5( = dxx )1( = cx
x
2
2
Example: dtt
t
)3(
92 = dt
t
tt
)3(
)3)(3( = dtt )3( = ct
t3
2
2
e. The Exponential Rule
The integral of the exponential function is the exponential
function. So, the integral of e
raised to the x is e raised to the x.
i.e. ex dx = ex + c
f. The Logarithmic Rule
When doing the power rule, we specified that it cannot be used
if the power is -1. But
why? Lets try it:
x-1 dx = )11(
)11(x =
0
0x =
0
1
So what do we do now????
Well if you recall: y = ln x implies dy/dx = 1/x = x-1
Since we suggested that integration is the inverse of
differentiation:
x-1 dx = 1/x dx = ln x + c
NOTE WELL:
3x4+ 3x2+3x dx
= 3 x4+x +x dx
BUT
3x x3 + x dx
Similarly
5x4 + x dx
5 x4 + x dx
Sample problems
1. 10x +2 dx
2. 3
122
x
xxdx
3. (y-2)(y+1) dy
Answers:
1. 5x2 +2x + c
2. x2/2 +4x+ c 3. y3/3y2/2 -2y+ c
Remember this as we will use it again
below.
The integral of x-1 is
ln x.
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g. Trigonometric Integration
When you integrate a trig function, you should always get
another trig function.
sin d = - cos + c
cos d = sin + c
tan d = -ln (cos ) + c
sec2 d = tan + c
h. Composite Functions
A composite function is one that has a function embedded in
another one.
Composite Power Functions
To integrate anything to a power (except -1) you add one to the
power and divide by the
new power (Power Rule) but for composite power functions you
must also differentiate
the portion enclosed in the brackets and then divide your
original answer by your
differentiation result.
For example: (3x2 + 5x)3 dx = {)13(
)53( 132 xx divided by (6x +5) } + c
= )56(4
)53( 42
x
xx + c
= 2024
)53( 42
x
xx + c
Composite Exponential Functions
Similarly, though the integral of e raised to any thing is e to
that thing, you must also
divide by the result of differentiating the thing that e is
raised to.
So, e5x dx = 5
5 xe + c
Composite Logarithmic Functions
Similarly, the integral of any thing raised to the power of
minus one is ln of the thing.
For a composite function raised to minus one, you must also
divide by the result of
differentiating the portion in brackets.
Example: dtt )35(
1
2 = (5t2+3)-1 dt
= t
t
10
)35ln( 2+ c
For example:
Function 1 . x2
Function 2 . 3x + 5
Putting Function 2
into Function 1 creates
a composite function
of (3x + 5)2
Sample problems
1. (7t3+3t) dt
2. (4y + 3)3 dy
3. 3xe dx
Answers:
1. ct
tt
963
)37(2
2
2
3
3
2. cy
16
)34( 4
3. cx
e x
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Remember to use all
the principles you
have learnt. Be careful.
If you cant do these
questions, review the
laws and try again.
Power Rule
Result of differentiating the brackets.
5 is the result of differentiating 5x w.r.t. x
10t is the result of differentiating (5t2+3) w.r.t. t
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Composite Trigonometric Functions
We can extend the concept to trigonometric functions.
The integral of sine of any thing is minus cosine of the thing,
but you must differentiate
the thing and divide by it.
Likewise, the integral of cosine of any thing is sine of the
thing, but you must
differentiate the thing and divide by it.
The integral of tangent and secant squared follows the same
principle.
So, sin (5 +4) d = 5
)45cos( + c
And, cos (10x2 + 5x) dx = 520
)510sin( 2
x
xx + c
Note: As you would have realized by now, it is important to
remember your rules of
differentiation as well. If necessary, review the
Differentiation self help book. Consider
the example below.
Example: e sin (3) d = { e sin (3) divided by d
d )3(sin(} + c
= { e sin (3) divided by 3cos(3) } + c
= )3cos(3
)3sin(e + c
Determining the Exact Original Function
Okay. So you now know the basic rules of Integration. See how
far you have come? But I
know you must be wondering Is there any way to know what c is?
Of course there is!
We said that an arbitrary constant (+c) had to be added to an
anti-derivate, since we do not
know which was the precise original function. But, if some
values of x and y for the
original function are given (i.e. we are given some boundary
conditions), we can actually
determine the exact original function. The best way to
illustrate this is to do a problem.
Consider the physics question below:
QUESTION 1: Find the equation describing the motion of an object
moving along a
straight line (i.e. an equation for x) if the equation for its
acceleration is given by a = 4t
2. The following is also known about the motion: At t = 5
seconds it velocity is 25 m/s. At
t = 12 s, the object has traveled 238 m from the origin.
More Problems
1. (cos(7t2) +sin(3-5t) dt
2. tan(5y) dy
3. xesin dx
Answers:
1. [sin(7t2)]/14t +
[cos(3-5t)]/5
2. c
y
y
2
1
2
1
5
)]5ln[cos(2
3. cx
e x
cos
sin
We can determine c
Here we get a Physics
based question. This
question shows you
exactly how
integration relates to
Kinematics.
Remember that
Velocity = a dt
Displacement = v dt
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6 a. We start with the fact that velocity, v = a dt
So v = 4t 2 dt
Applying the Power Rule and the Constant Rule
v = 2t2 2t + c
Now let us also use the other information we know to solve for
c
i.e. at t = 5 s, v = 25 m/s
25 = 2(5)2 2(5) + c
-15 = c
Therefore v = 2t2 2t 15
b. But we want a formula for displacement, x.
We utilize the fact that x = v dt
= 2t2 2t 15 dt
= 33
2t - t2 - 15t + q
Now let us also use the other information we know to solve for
q
i.e. at t = 12 s, x = 238 m
238 = 2(12)3/3 (12)2 15(12) + q
-590 = q
Therefore x = 33
2t - t2 - 15t - 590
Wasnt that easy? You try the question below.
QUESTION 2: The rate of change of resistance (R) with respect to
temperature (T) of
an electrical resistor is given by dR/dT = 0.009T2 + 0.02T 0.7.
Find the resistance when
the temperature is 30oC if when R = 0.2 when T = 0oC.
Well start you off:
Since dR/dT = 0.009T2 + 0.02T 0.7
dR = 0.009T2 + 0.02T 0.7 dT
dR = 0.009T2 + 0.02T 0.7 dT
R =
[See if you get R = 88.1 for your answer].
Any letter can be
used for the arbitrary
constant. In this case,
to avoid confusion
between the different
parts of the question,
we use +q for the
displacement
equation
This is another
Physics application of
integration.
HINTS FOR
SOLVING
a. Find the
equation for R
by integrating.
b. Substitute for T
and R to
determine c in
the equation.
c. Find R at T =
30oC
We are representing the constant
here by q
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Definite Integrals
So we are almost at the end. One last concept.
Integration was developed as a way to find the sum of a number
of quantities. When this is
being done you are finding a definite integral and your final
answer is a numerical value.
This is achieved by integrating between limits.
So, a definite integral is denoted by its limits:
itupper
itlower
lim
lim
As an example:
3
1
23 dxx means integrate 3x2 w.r.t. x between the lower limit x
=1
and the upper limit x = 3.
So how do we do a definite integral problem. Simple. We follow
two steps.
Step 1: Treat the integral as an indefinite one and do the
integration but leave off
the constant c.
Step 2: Now substitute the upper limit in the result from Step 1
and subtract the
result of substituting the lower limit in the result from Step
1.
Okay, lets work an example.
Evaluate
2
0
22 )5( dttt = )110(3
)5( 322
0t
tt
= )1)2(10(3
)2)2(5 32 minus
)1)0(10(3
)0)0(5 32
= 57
5832 - 0
= 102.3
Conclusion
Now you know how to integrate (at least the basics). In your
upcoming physics lectures,
you will learn when to use it for physics and where it applies.
If you think about it youve
come a long way from where you started. But all of this will
hard work will be wasted if
you do not practice to differentiate. PRACTICE, PRACTICE,
PRACTICE. Its the only
way to keep integration fresh in your mind.
Isnt it nice that it
also ended up as the
opposite of
differentiation??!!!??
Sample problems
1.
5
2
2x dx
2.
6
2
3 dx
3.
2
3/
sin x dx
4.
3
1
2 )14( xx dx
Answers:
1. 19 2. 12 3. -1/2 4. -16/3 Also try this
question:
A Force moves an
object from x = 0m to
x = 3m according to
F = x3 x. Find the
Work Done.
Remenber W = F.dx
(Ans: 15.75 Joules)
Substitute upper limit
Substitute lower limit
Result from Step 1
