Basic Integration
Integrating can also be defined as “finding the anti-derivative.”
***Derivatives and Integrals are the inverses of each other***
An equation is derived when it goes from f(x) to f’(x)
An equation is integrated when it goes from f’(x) back to f(x)
Symbol of Integration is:  f (x)
Basic Rules
 0dx = c

 kdx = x+ c

xn = xn+1/ n+1 + c
1
 x dx = ln|x| + c
exdx = ex + c

 f ( x)  g ( x)dx =  f ( x)dx +  g ( x)dx
uxdx = ux/ ln u + c
****BE SURE TO ADD “+ C” TO YOUR FINAL ANSWER****
ALL YOU DO FOR POLYNOMIALS IS GO UP A POWER AND DIVIDE BY THE NEW POWER!!!
Easy Examples:
1.
2.
3.



x2 + x – 2dx
x3 + 3x2 + 5xdx
5xdx
Answers to Examples
1. X3/3 + x2/2 - 2x + c
2. X4/4 + x3 + 5x2/2 + c
3. 5x/ln 5 + c
Properties of Definite Integrals
***DO NOT ADD “ + C” TO YOUR ANSWERS***
b
Definite integral (Riemann integral):
 f ( x)dx
a
a
Zero:
 f ( x)dx  0
a
Order of Integration:
Constant Multiple:
b
a
a
b
 f ( x)dx   f ( x)dx
b
b
a
a
 kf ( x)dx  k  f ( x)dx
Sums and Differences:
b
b
b
[ f ( x)  g ( x)]dx   f ( x)dx   g ( x)dx
a
a
Additivity:
a
b
c
c
a
b
a
 f ( x)dx   f ( x)dx   f ( x)dx
Simplify these Problems:
1.
4
5
1
4
 5x   5x 
10
2.
 4x  3 
9
4
3.

x2+9𝑥 + 12 =
4
11
4.
 4(3x  3) 
8
OR
b
b
b
a
a
a
[ f ( x)  g ( x)]dx   f ( x)dx   g ( x)dx
U
SUBSTITUTION
Sometimes, general rules do not enable us to integrate. This is where USubstitution comes in. In the words of Dr. Stotler, U, USUALLLYYY replaces
the term(s) inside parenthesis, a radical, or a trig function.
Steps:
1. Given f(g(x)); Allow u= g(x)
2. Differentiate: du= g’(x) dx
3. Rewrite the integral in terms of U
4. Evaluate the integral
5. Replace the u with the original g(x)
6. Check you result [Take the derivate of your final answer]
u=
du
=
dx
du =
** Let u = g(x) and du= g’(x) dx
Algebraic Functions
Example 1 :
 x(x  1)
10
1. Let u= x+1, Then x= u-1
2. Differentiate: du = dx
10
(
u

1
)
u
du

3. Rewrite:
=  (u 11  u 10 )du
u 12 u 11

c
12 11
4. Integral:
( x  1) 12 ( x  1) 11

c
5. Replace u:
12
11
TRY PLEASE
1)
2)
 x x  2dx
 2 x  5 dx
2/3
Example 2 :
 x
x2
3
8

5
dx
1. Let u= x3-8
1
2. Differentiate: 3du= x2 dx
3. Rewrite
1 5
u du
3
1  u 4 
c
4. Integrate: 
3   4 
5. Replace u:

1
x3 8
 12

4
c
Inverse Trigonometric Functions
Example 1 :

dx
9  4x 2
1. Let u2= 4x2, so u=2x
2. Let a2=9, so a=3
1
3. Differentiate: du= dx
2
4. Rewrite:
1
du

2
a2 u2
1
𝑢
sin−1 (𝑎)+c
2
1
2𝑥
6. Replace u: sin−1 ( 3 )+c
2
5. Integrate:
Logarithmic and Exponential Functions
Example 1 :

x2  3
dx
x 1
1. Use division[(x2+3) by (x-1)] to re-write the problem as x+1+
4
𝑥−1
2. Let u= x-1
3. Differentiate: du= dx
1
4. Rewrite: 4 du
u
5. Integrate: 4 ln u +c
6. Replace u: 4 ln x 1 +c
7. Integrate: x+1
8. Put it all together:
x2
 x  4 x 1  c
2
Example 2 :
 sin x e
cos x
dx
1. Let u=cos x
2. Differentiate: -  sin x dx; 
3. Rewrite:

1
e

u
du

 sin xdx
du
4. Replace u and integrate:

1

e cos x  c
Trig and Inverse Trig Integrals
Trig Integrals
 sin( x)dx = -cos(x)

+ c
csc2(x)dx = -cot(x) + c
 cos( x)dx = sin(x) + c

 sec( x) tan( x)dx = sec(x) + c
 csc( x) cot( x)dx = -csc(x) + c
sec2(x)dx = tan(x) + c
Inverse Trig Integrals

du/ (a2-u2)1/2 = sin-1(u/a) + c

du/ u(u2-a2)1/2 = (1/a)sec-1(u/a) + c

du/a2 + u2 = (1/a)tan-1(u/a) + c
****SEE THE YELLOW SHEET FOR MORE SPECIFIC INTEGRATION FORMULAS****
Practice Problems with Trig/ Inverse Trig using U-Sub
1.
2.
3.
4.




dx / 1 + 9x2
dx / (1-4x2)1/2
dx / 2x(4x2-1)1/2
tanxdx
Answers to Problems
1. u2 = 9x2
3. u2= 4x2
u=3x
2. u2= 4x2
du= 3dx
1
du= dx
3
1
3
du / 1 + u
1
tan-1u + c
3
1
tan-1(3x) + c
3
2
u= 2x
u= 2x
du = 2dx
du= 2dx
1
du = dx
2
1
du= dx
2
1
du / (1-u2)1/2
2
1 -1
sin (2x) + c
2
1
du / u (u2 – 1)1/2
2
1
sec-1 (2x) + c
2
4.  tan xdx =  sin xdx / cos x
u=cosx
du= -sinxdx
-  du / u
-ln|u| + c
-ln |cosx| + c
The Fundamental Theorem of
Calculus
The fundamental theorem of calculus has two parts:
Part One: Deals with definite integrals
 Definite Integrals
o Have boundaries called the limits of integration
o Have numeric answers
o Used to find exact area underneath a curve
Let f be a real-valued integrable function defined on a closed interval [a, b]. If F is a function
such that F ′(x) = f(x) for all x in [a, b] (that is, F is an antiderivative of f), then
In plain terms, when given boundaries on the integral you can
1. Integrate
2. Substitute the top number into the integrated function
3. Substitute the bottom number into the integrated function
4. Subtract the entire second integrated/substituted number from the first to
determine the numerical value
5. Don’t worry about the +C!
𝟓
Ex 1: ∫𝟏 (𝒙𝟐 + 𝟐𝒙 − 𝟏)𝒅𝒙 =
Integrate:
𝑥3
3
+ 𝑥 2 − 𝑥|5
1
Substitute: [125/3 + 25 – 5] – [1/3+1-1]
Solve: ≈ 61.333
𝟔
Ex 2: ∫𝟐 𝒙𝟐 (𝟔𝒙𝟑 + 𝟒)𝒅𝒙 =
Integrate using U-sub: u= 6𝑥 3 + 4 du= 18𝑥 2 𝑑𝑥
6
(6𝑥 3 +4)2
|
2
36
Substitute: [1690000/36] – [ 2704/36]
Solve: 46944.444 – 75.1111 ≈ 46869.33
𝑑𝑢
18
6
1
= 𝑥 2 𝑑𝑥 ∫2 𝑢𝑑𝑢 = (18)
𝑢2
2
𝑑𝑢=
Definite Integrals with U-sub:
6
Use the example from before ∫ 𝑥 2 (6𝑥 3 + 4)𝑑𝑥 =
2
1 𝑢2
Integrate: ( )
𝑑𝑢|6
18 2
2
Substitute the boundaries into the equation for u: u= 6𝑥 3 + 4
6(6^3)+4 = 1300
6(2^3) + 4 = 52
These are your new boundaries!
1300
∫52
1
(18)
𝑢2
2
𝑑𝑢
1
Now substitute: [(18)
13002
2
1
] – [(18)
522
2
]
Solve: [46944.444] – [75.1111] ≈ 46869.33
𝜋/2 𝑐𝑜𝑠𝑥
Ex 2: ∫0
+1=1
2 𝑑𝑢
∫1
𝑢
𝑠𝑖𝑛𝑥+1
𝑑𝑥
u = sinx + 1 du = cosx New boundaries= sin(𝜋/2) + 1 = 2 & sin(0)
= ln(u) = ln2 – ln1 = .6931
Definite Integrals on the Calculator:
Yes you can integrate on your calculator! It will not integrate indefinite integrals, but it will
integrate definite integrals.
1.
2.
3.
4.
Math button
9
Plug in the boundaries, the equation, and the variable of integration
Voila! You have the answer!
What’s the point?
Consider the problem of finding the area ‘A’ between a curved line: y = f(x) and the ‘x’ axis,
between x = a and x = b:
The fundamental theorem of calculus allows you to add up all the extremely tiny rectangles
that you can draw underneath the curve. See Riemann Sums.
Evaluate without a calculator:
𝜋/2
1. ∫–𝜋/2 𝑐𝑜𝑠𝑥 𝑑𝑥
9
2. ∫1 2𝑥(𝑥 2 + 4) 𝑑𝑥
1
3. ∫0 (𝑥 4 − 5𝑥 3 + 3𝑥 2 − 4𝑥 − 6)𝑑𝑥
Evaluate with a calculator:
9
4. ∫1 2𝑥 √𝑥 𝑑𝑥
Answers:
1.
2.
3.
4.
5.
2
3600
-161/20
193.6
16
4
5. ∫−4|𝑥| 𝑑𝑥
Part Two: Deals with the derivative of a definite integral
𝑥
If F(x) is continuous on [a,b] then the derivative of function 𝐹(𝑥) = ∫𝑎 𝑓(𝑡) is:
𝑥
𝑑𝐹
𝑑
=
∫ 𝑓(𝑡)𝑑𝑡 = 𝑓(𝑥)
𝑑𝑥
𝑑𝑥
𝑎
Looks like a lot, but it is really easy. In an integral where the boundaries are a (a constant)
to x (a variable) the derivative of said integral is just the equation with x substituted for t.
𝑑
𝑥
Example 1: 𝑑𝑥 ∫−1 1 + 𝑡 2 𝑑𝑡 = 1 + 𝑥 2
Any constant on the bottom will give the same answer.
𝑑
𝑥
𝑑
𝑥
Example 2: 𝑑𝑥 ∫12 1 + 𝑡 2 𝑑𝑡 = 1 + 𝑥 2
Example 3: 𝑑𝑥 ∫19670 1 + 𝑡 2 𝑑𝑡 = 1 + 𝑥 2
*** If the upper limit is a function of x instead of just plain x you must multiply f(x) by the
derivative of x
𝑑
𝑥2
𝑑
𝑥
𝑑
𝑥 2 +2𝑥 1
Example 4: 𝑑𝑥 ∫−1 1 + 𝑡 2 𝑑𝑡 = (1 + 𝑥 2 )(2𝑥) = 2𝑥 + 2𝑥 3
Try:
𝑑
1. Find 𝑑𝑥 ∫1 𝑠𝑖𝑛𝑡 𝑑𝑡
3. Find
∫
𝑑𝑥 1
Answers:
1. Sinx
2. 3(𝑥 2 -x)
3.
2𝑥+2
(𝑥−2)
𝑡−2
3𝑥
2. Find 𝑑𝑥 ∫5 (𝑡 2 − 𝑡)𝑑𝑡
𝑑𝑡
1
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