Misconceptions Of Mathematics Techniques 2008 Paper
1. Many students wrote
 xe
x
dx   x dx  e x dx . This is a question on
integrating by parts and you cannot carry out the above operation:
x
x
 xe dx   x dx  e dx  Not Equal
Some students did not carry out the second integration. Although you
get the same answer, that is
x
x
x
 xe dx  xe  e  C
should be
 xe
x
dx  xe x   e x dx and then evaluate integral of e x .
2. Another misconception on integration was the following:
dx
1
1
  x  1 x  2    x  1  x  2 dx
and then they use integration by parts with u   x  1
1
and dv   x  2  .
This is a question on integration by partial fractions. You need to
1
convert the integrand
into partial fractions and then
 x  1 x  2
integrate.
3. The main misconception on second order differential equations was
when students try to find the complementary function and some
students carried out the following derivation:
m2   2  0
Step 1
m 2   2
m   ,
m 
Step 2
Step 3
Step 3 in wrong because when we solve m2   2 we get m   j .
Remember we are taking the square root of a negative number.
4. Another error that students made was the following
1
1
sin  x 
1
dx

sin  x  dx
0 x
x 0
You cannot do the above operation unless x is a constant.
5. Some students also made some elementary arithmetic mistakes
such as
182   6   12  324  36  1
2
They should know  6   36 not  36 . Some of them wrote  6  as
2
6 2 .
2
6. Most students managed to find the first derivative of the
parametric equations. The correct answer for this part was
cos  3t 
dy
 3
but no one managed to get the correct answer for
dx
sin  t 
d2y
second derivative
. They applied the quotient rule on
dx 2
cos  3t 
dy
d  cos  3t  
 3
that is they found
 3
 . THIS IS
dx
sin  t 
dt 
sin  t  
INCORRECT because
d  dy 
d y dt  dx 

dx 2
dx / dt
2
7. Many students forgot to add the constant C in the evaluation of
indefinite integrals.
8. Another common error was the following:
2
d2 4
d
 x  2 x 2  1   x 4  2 x 2  1 
2 


dx
dx 
d2
Remember the notation
means find the second derivative of the
dx 2
function and not square it.