# lesson 44--other antiderivatives and particular solution ```Aim: How do we find anti-derivatives of elementary functions? 2-day lesson
Objectives: to find the anti-derivatives of trig, exponential and inverse trig functions, to find the
particular solution, to find the anti-derivative of a product or quotient of polynomials
Grouping: students are given opportunity to work in cooperative setting during this class. Time is
purposely set aside for students to work collaboratively on at least one set of exercise problems. During
this time, students are encouraged to explain concepts and materials and solutions to each other.
Furthermore, students get to present their solutions on the board.
Differentiated instruction: all students are held to the highest standard. However the students that are not
performing well are grouped with students that are excelling to serve as study partners. Teacher will
informally assess their understanding by asking them questions to ensure that they are keeping up.
Assessment: Class is given multiple occasions to work individually and in groups. Teacher circulates the
room to assist and also assess student understanding. Furthermore, students are encouraged to explain
work to each other. This is another opportunity for teacher to assess student understanding. Lastly, when
lesson is finished, students are grouped for class work, questions that are designed as exit slip problems.
Homework Review (10 minutes): Students are assigned problems to present on board. Students will also
answer questions. Teacher can go over a problem if no one in class could explain it
Lesson Development: [15 – 20 min]
Just as there are several ways to mean differentiation—differentiate, taking the derivative, or finding the
instantaneous rate of change, there are several ways to mean integration—indefinite integral, antiderivative, or solve the differential equation*
Example: What does solving the differential equations mean? We know that  asks us to find the
function when given its derivative function. “Solve the differential equation” means the same except
sometimes it comes with an initial condition which enables us to find the value of c.
Example:
a) f '( x)  4 x, f (0)  6
b) h '(t )  8t 3  5, h(1) = -4
x2
f ( x)   4 xdx  4[ ]  c  2 x 2  c
2
2
f (0)  2(0)  c  6
c6
t4
h(t )   (8t  5)dt  8( )  5t  c
4
4
h(1)  2(1)  5(1)  c  4
c  11
Particular Solution: f ( x)  2 x 2  6
Particular Solution: h(t )  2t 4  5t  11
3
c) f ''( x)  x 3/2 , f '(4)  2, f (0)  0
f '( x)   x 3/2 dx 
x 1/2
c
(1/ 2)
f '(4)  2(4) 1/2  c  2, c = 3
f '( x)  2( x) 1/2  3
x1/2
f ( x)   (2 x  3)dx  2(
)  3x  c
1/ 2
f (0)  4(0)1/2  3(0)  c  0, c = 0
1/2
Particular Solution: f ( x)  4 x1/2  3 x
d) f ''( x)  sin x, f '(0)  1, f (0)  6
f '( x)   sin xdx   cos x  c
f '(0)   cos 0  c  1, c  2
f '( x)   cos x  2
f ( x)   ( cos x  2)dx   sin x  2 x  c
f (0)   sin 0  2(0)  c  6, c = 6
Particular Solution: f ( x)   sin x  2 x  6
EX1: An evergreen nursery usually sells a certain shrub after 6 years of growth and shaping. The growth
dh
 1.5t  5 , where t is the time in years and h is the height
rate during those 6 years is approximated by
dt
in centimeters. The seedling are 12 centimeters tall when planted (t=0). [10 – 12 min]
a) Find the height after t years
t2
h   (1.5t  5)dt  1.5  5t  c
2
To solve for c, we plug the initial condition (0, 12)
back to the equation we just found:
1.5 2
12 
(0)  5(0)  c
2
c  12
h  1.5
b) how tall are the shrubs when they are sold?
The shrubs are sold after 6 years, so t = 6.
h  1.5
(6) 2
 5(6)  12  1.5(18)  42  69cm
2
t2
 5t  12
2
Going forward, we will use the dominant terms--indefinite integral, anti-derivative--interchangeably. [8
– 10 min]
We know that:
d
d
d
sin x  cos x,
cos x   sin x,
tan x  sec 2 x,
dx
dx
dx
d x
d
1 d
d
e  ex ,
ln x  , a x  ln a( a x ),
sec x  sec x tan x
dx
dx
x dx
dx
d
1
d
1
tan 1 x  2 ,
sin 1 x 
dx
x  1 dx
1  x2
 sin xdx   cos x  c,  cos xdx  sin x  c,  sec
Therefore:
x
x
 e dx  e  c,
x
a 
ax
,
ln a
2
1
 x dx  ln x  c,  x
 sec x tan xdx  sec x  c, 
1
1  x2
EX1: If f '( x)  e x  3 and f (0)  2 , what is
f ( x) ? [5 min]
xdx  tan x  c
1
dx  tan 1 x  c
1
2
dx  arcsin x  c
EX2: If f '( x) 
3
 x and f (1)  2 , what is
x
f ( x) ? [5 min]
3
x2
 xdx  3ln x   c
x
2
1
f (1)  3ln1   c  2
2
3
c
2
x2 3
f ( x)  3ln x  
2 2
f ( x)   e x  3dx  e x  3x  c
f ( x)  
f (0)  e0  c  2
c 1
f ( x)  e x  3x  1
Is  ( x  1)( x  2)dx 
  ( x 1)dx   ( x  2)dx  ? Verify it using wolframalpha.com
How can we integrate this? Since we know  f ( x)  g ( x)dx   f ( x)dx   g ( x)dx , why don’t we turn
this product into a sum or difference? How?
2
 ( x  1)( x  2)dx   ( x  x  2)dx 
x3 x 2
  2x  c
3 2
Same is true if we are asked to integrate a quotient.
x3  2 x  1
1
x3
2
 x dx   x  2  x dx  3  2 x  ln x  c
EX3: Integrate [10 – 12 min]
1 y2
a) 
dy
y

b)  ( x  2) 2 dx
1 y
1
y
dy    ydy  ln y   c
y
y
2
2
2
2
2
 ( x  2) dx   x  4 x  4dx 
x3
 2x2  4x  c
3
c)


2 x5 8 3/2
x ( x  2)dx   x3/2  2 x1/2 dx 
 x c
5 3
x ( x  2)dx

d)
( x  2)
dx
x
  x1/2  2 x 1/2 dx 
2 x3/2
 x1/2  c
3
HW#44: P291 – 293: 23 – 28, 31, 32, 34 – 37, 39, 41 – 44, 63, 66, 69, 70, 74 (due in two classes since
this is a 2-day lesson)
HW#44 Solutions: P291 – 293: 23 – 28, 31, 32, 34 – 37, 39, 41 – 44, 63, 66, 69, 70, 74
23)
24)
25)
26)
27)
28)
31)
32)
34)
36)
39)
35)
37)
41)
42)
44)
43)
63)
69)
66)
70)
74)
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