Integration
Area and Volume
Session 1
Integration
Tips on integrating
Areas between curves
Volumes of Solids of Revolution
Exam questions
Idea behind Integration
Each rectangle is f(x) high
and dx wide so…
Idea behind Integration
Each rectangle is f(x) high
and dx wide so…
Area under the curve is
all these rectangles
added together or…
Idea behind Integration
Each rectangle is f(x) high
and dx wide so…
Area under the curve is
all these rectangles
added together or…
Integral is limiting sum of
these rectangles as
dx goes to zero…ie
Another way to look at primitives
past
present
f(x
)
is
fun
cti
on
va
lue
future
Another way to look at primitives
past
present
f(x
)
is
fun
cti
on
va
future
f’(x) is gradient of tangent
lue
Another way to look at primitives
past
present
f(x
)
Gives area under curve
is
fun
cti
on
va
lue
future
Fundamental Theorem of Calculus
The area enclosed by the curve y = f(x), the x-axis
and the lines x = a and x = b is given by
Where F(x) is the primitive function of f(x).
Some tips when integrating
• indefinite integrals.. don’t forget to add c
Some tips when integrating
• indefinite integrals.. don’t forget to add c
• definite integrals.. F(a)-F(b) remember the minus
Some tips when integrating
• indefinite integrals.. don’t forget to add c
• definite integrals.. F(a)-F(b) remember the minus
• do not differentiate when integrating, we find the
primitive !
Some tips when integrating
• indefinite integrals.. don’t forget to add c
• definite integrals.. F(a)-F(b) remember the minus
• do not differentiate when integrating, we find the
primitive !
• And remember logs when all else fails
Some tips when integrating
• No product or quotient rule
• Always check you result by differentiating it
Some tips when integrating
• Simplify before integrating
Eg:
2010 Question 2
(try these now!)
120 marks in 180minutes these should take 6 x 1.5 = 9 minutes
Solutions
Notes from the marking centre
d) (i) Most candidates knew that. A
significant number of candidates multiplied
by 5 rather than dividing and some added
5 to in the denominator. Some candidates
used some form of differentiation
instead of integration and many forgot to
include the constant of integration in
their final answer.
Notes from the marking centre
(ii) Most candidates realised that a logarithm integration was needed.
Most incorrect answers had a term involved but did not correctly deal
with the coefficient or with the numerator. Again, some differentiated
while others thought there was an inverse tan integral involved. Of
concern was a technique that used the expression outside the integral
sign as it incorrectly moves the variable outside the integral.
(e) Many different correct solutions were in
evidence. Candidates who found incorrect
primitives had difficulty gaining any marks.
A novel solution was to draw a sketch
of y = x + k from 0 to 6 and equate an
expression for the area of the trapezium
so formed to 30.
Solutions
Area between Curves
Area between Curves
Area between two curves
Find the green area…try this now!
Solution- just subtract functions and integrate
Area between three lines…plain sailing
try this now!
Solutions
“The majority of
candidates knew to use
a definite integral
involving the difference
of the functions”….
Solutions
“Common errors
included incorrect limits,
incorrect primitives,
addition (rather than
subtraction) of the
functions and subtraction
of the functions in the
wrong order.”
Trig can also be integrated
…try these now!
(2010 Question 5)
solutions
Put in terms of sin/cos
Left Hand Side is
solutions
Left Hand Side is
solutions
Left Hand Side is
Difference of squares
= RHS as required
Solutions
Area between curves
…a bit harder ..try this…
(2010 Question 5)
Solutions
This part was generally well done
with many candidates able to earn at
least one mark. A common error,
however, was to state that the sum
of the areas was 1 and not the
correct value of 2. It was noticeable
that a number of candidates were
unable to solve the log equations
by simply writing the expressions in
exponential form and therefore
missed an easy third mark.
Solutions
This part was generally well done
with many candidates able to earn at
least one mark. A common error,
however, was to state that the sum
of the areas was 1 and not the
correct value of 2. It was noticeable
that a number of candidates were
unable to solve the log equations
by simply writing the expressions in
exponential form and therefore
missed an easy third mark.
So read and reread the question and learn your logs
2009 Question 2
(i) Find
mark)
(1
(ii) Find
marks)
(2
(ii) Evaluate
(3 marks)
Solutions
Candidates who showed
setting out which included the
first line of working generally
gave a final response that
achieved full or part marks.
Solutions
Candidates were more successful when
they explicitly showed the substitution of
their limits.
Volume of Solids of Revolution
Around x axis so y = f(x)
Around y axis so x = g(y)
Volume of Solids of Revolution
Volume of revolution
(2009 Question 6) try this now!
Solutions
More variety!
(2008 Question 3&5) (7 marks=10.5 minutes)
2008 Question 6
6(c)
solutions
3(b) (i)
3(b) (ii)
solutions
Q5 (a)
solutions
Q5 (a)
solutions
Q5 (a)
solutions
Q5 (a)
References
• www.boardofstudies.nsw.edu.au
• Look at notes from the marking centre.