Groundwater Lab Report/ANSWER SHEET
Answer the lab questions for this week and summarize the lab experience using this
Pay special attention to the graphs and figures.
Complete this weeks lab by filling in your responses to the questions from the
1201SCE MATHS 1A
Problem & Solutions sheet for week 4
These questions are based the week 3 lectures
Question 1 Practice question on determinants
Calculate the following determinants and check that you get the answers given
1201SCE problem sheet for w
4.4 THE DEFINITE INTEGRAL
For any given value of t, the area is given by the limit of the
where for each i, ci is taken to be any
point in the subinterval [xi 1, xi].
For any function f defined on [a, b], the definite integral
4.5 THE FUNDAMENTAL THEOREM OF CALCULUS
THEOREM 5.1 (The Fundamental Theorem of Calculus, Part I)
If f is continuous on [a, b] and F(x) is any anti derivative of f(x),
( ) = ( ) ( )
We will often use the notation
( )| = ( ) ( )
This enables us to w
5.2 VOLUME: SLICING, DISKS AND WASHERS
Any solid whose cross sections perpendicular to some axis
running through the solid are all the same. We call any such
solid a cylinder
The volume of a right circular cylinder is
while the volume of a box is
6.3 TRIGONOMETRIC TECHNIQUES OF INTEGRATION
Integrals Involving Powers of Trigonometric Functions
Consider integrals of the form
Case 1: m or n Is an Odd Positive Integer
1. Evaluate cos4x sinx dx
2. Evaluate cos4x sin3x dx.
3. Evaluate 5
Case 2: m an
6.4 INTEGRATION OF RATIONAL FUNCTIONS USING PARTIAL
In this section, we introduce a method for rewriting certain rational
functions that is very useful in integration as well as in other
applications. We begin with a simple observation. Note tha
6.2 INTEGRATION BY PARTS
can not be evaluated with what you presently
INTEGRATION BY PARTS
1.Evaluate x sinxdx
A Poor Choice of u and dv
2. Evaluate lnx dx.
3. Evaluate x2sinxdx.
4. Evaluate e2x sinxdx.
5.3 VOLUMES BY CYLINDRICAL SHELLS
In this section, we present an alternative to the method of
washers discussed in section 5.2. Let R denote the region
bounded by the graph of y = f(x) and the x-axis on the interval
[a, b], where 0 < a < b and f(x) 0 on [
Applications of the Definite Integral
5.1 AREA BETWEEN CURVES
1. Find the area bounded by the graphs
of y = 3 x and y = x2 9.
2. Find the area bounded by the graphs of y
= x2 and y = 2 x2 for 0 x 2.
A Case Where the Intersection Points Are Know
4.8 THE NATURAL LOGARITHM AS AN INTEGRAL
For x > 0, we define the natural logarithm function, written ln x,
Graph of =
For any real numbers a, b > 0 and any rational number r,
(i) ln1 = 0
(ii) ln(ab) = lna + lnb
4.1 ANTI DERIVATIVES
Finding a way to Undo the derivative
1. Find an anti - derivative of () = 2
Suppose that F and G are both anti - derivatives of f on an
interval I. Then, ( ) = ( ) + , for some constant c.
4.6 INTEGRATION BY SUBSTITUTION
2. ( 3 + 5)100 (3 2 )
INTEGRATION BY SUBSTITUTION
Integration by substitution consists of the following general steps,
as illustrated in example 2.
Choose a new variable u: a common choice is the innermost
3.3 MAXIMUM AND MINIMUM VALUES
For a function f defined on a set S of real numbers and a number c S,
(i) f(c) is the absolute maximum of f on S if f(c) f(x) for all x S and
(ii) f(c) is the absolute minimum of f on S if f(c) f(x) for all x
3.5 CONCAVITY AND THE SECOND DERIVATIVE TEST
(i) concave up on I if is increasing on I or
(ii) concave down on I if is decreasing on I.
For a function f that is differentiable on an interval I, the graph of f is
Concave up, increasing
Name: . Index No: .
Date: . Signature.
MOKASA JOINT EVALUATION EXAMINATION
Kenya Certificate of Secondary Education
INSTRUCTIONS TO CANDIDATES
Write your name, index number, signature and date in the
NAMEINDEX NOADM NO.
MOKASA JOINT EXAMINATIONS
Kenya Certificate of Secondary Education (K.C.S.E)
TIME: 2 HOURS
INSTRUCTIONS TO CANDIDATES.
Write your name and ind
QIBT 1201SCE , Mathematics 1A,
Solutions for Problem Class 2 Due week 8
The range of a function f ( x) = the x values where the function f ( x) is sensibly
The domain of a function f ( x) = the possible values that the function can tak
Problem Class 1 - Solutions
In this problem you need to look at the
relative sizes of the vectors and also at
The lines AB and CD are parallel and have the same length. But the direction
from A to B is opposite to the dire
1201SCE MATHS 1A
Problem sheet & answers for week 2
All the questions for this week are key questions
World cigarette production increased approximately linearly between 1970 and 1990 In
1970, production was 3 trillion and in 1990 it was 5
1201SCE MATHS 1A
Problem sheet for week 3
*Question 2 Adding velocities (section 6.2, p 39-41)
In the diagram below a boat starts out traveling on a lake at 30 km/hr at a direction of 30
degrees. (above the East direction). The velocity of the boat i
1201SCE/1602SCE Answers for midsemester exam 2005
Question 1 (a) The first few lines of the Pascal triangle are easy to write down. Then use the rule that new entry = value just above + value above and to left 1 1 1 1 1 2 3 4 (x + y)2 (x + y)3 (x + y)4
PROBLEM SHEET WEEK 10 SOLUTIONS.
Students should attempt questions 1(a),(f) and 2.
y = x 4 - 12x 3 + 48x 2 - 64x .
All values of x.
Intercepts on the x-axis and the y-axis.
When x = 0, y = 0. Therefore curve crosses at the origin.
When y =
Griffith University, School of Science 1201SCE Mathematics 1A and 1602SCE Mathematics Midsemester examination, April 2005 Time allowed Reading time Working time 10 minutes 2 hours 30 minutes.
READ THE FOLLOWING INSTRUCTIONS CAREFULLY This examination cont
Answers for midsemester exam Question 1
This section will often be answered in a complicated way there are 3 key points The horizontal line test says that y = f ( x) has an inverse function if a horizontal line meets the graph of this function at just
Q1. Determine whether the relation represents a function. If it is a function, state the domain and range.
domain:cfw_16, 20, 24, 28
range: cfw_4, 5, 6, 7
domain: cfw_4, 5, 6, 7
range: cfw_16, 20, 24, 28
c. not a function
38 X11423 XX27575X 100
2 X 19 X 23 X 3
X 10 2
2 X 3 X 19 X 5 2 X 23
3 2 X 10 2
y 52 x 2
M 1 52
M 2 52
10 2k 25
1 180( n 2 )
180n 360 1080