Mat h 12 06 Ca lc ul us – Sec . 4. 8: A ntid er ivat ives
I.
Revi e w fro m chapt er s 2 & 3 :
A. Derivative of
B. Notation:
f (x) : lim
h!0
dy
, y!, f !(x),
dx
C. Applications: a.
II.
f ( x + h) - f ( x )
h
df
,
dx
d
( f (x)) , Dx f
dx
f !(c) =slope of tangent line to graph of y = f (x) at the pt (c, f (c)) .
b.
f !(c) = the rate of change of f wrt x when x=c.
c.
v(c) = s!(c) =the rate of change of position wrt time when t=c.
A ntid er ivativ e s
A. Find the derivative of
1.
F(x) = x 2
2.
F !(x) =
Therefore, if
B.
F(x) = x 2 !
F(x) = x 2 + 7
3.
F !(x) =
F !(x) =
3
4
f (x) = 2x , then x 2 + C is the antiderivative of f (x)
D ef n : A function
F is an anti de riv ative of a function f on an interval I , if
F !(x) = f (x) ! x in I .
C. Th m : If
F is an antiderivative of f on an interval I , then the most general antiderivative
of f on I is F(x) + C where C is an arbitrary constant.
D. Use your knowledge of derivatives to find an antiderivative of the following.
1.
f (x) = 3x 2
2.
G(! ) =
F(x) =
III.
g(! ) = cos(! )
3.
h(y) = 2y
H (y) =
Inde fi nit e Inte gr al
A. De f n : The set of all antiderivatives of
x, denoted by
Integral sign
B.
! f (x) dx .
integrand
! f (x) dx
! f (x) dx = F ( x ) +C
f is the inde finit e i nt egr al of f with respect to
x is the variable of integration
iff F '( x ) =f ( x )
C. Any 2 antiderivatives of a function differ by only a constant.
D. An indefinite integral
b
! f (x) dx is a function (or a family of functions), while a definite
! f (x) dx is a number.
integral
a
E. Formulas:
k is a nonzero constant
1.
!k
2.
n
! x dx =
3.
! x dx = !
4.
! e dx
dx = kx + C
1
x ( n+1)
+ C ; n " Q , n # $1
n +1
dx
= ln x + C, x " 0
x
= ex + C
x
ax
+ C, a > 0, a " 1
5. ! a dx =
ln a
x
!e
kx
!a
dx =
kx
1
k
ekx + C
" 1 % kx
dx = $
a + C, a > 0, a ( 1
# k ln a '&
6.
" sin x dx = ! cos x + C
" sin(kx) dx = !
1
k
7.
! cos x dx = sin x + C
! cos(kx) dx =
sin(kx) + C
8.
! sec
x dx = tan x + C
! sec (kx) dx =
9.
" csc
x dx = ! cot x + C
" csc (kx) dx = !
10.
! sec x tan x dx = sec x + C
! sec(kx) tan(kx) dx =
11.
" csc x cot x dx = ! csc x + C
" csc(kx)cot(kx) dx = !
12.
"
13.
! 1+ x
14.
"x
2
2
dx
1! x
dx
2
2
= sin !1 x + C, x 2 < 1
= tan "1 x + C
dx
x !1
2
= sec !1 ( x ) + C, x 2 > 1
2
2
"
dx
1! k x
2
dx
! 1+ k
"x
2
x
2
tan(kx) + C
1
k
cot(kx) + C
1
k
sec(kx) + C
1
k
csc(kx) + C
1
k2
= 1k tan "1 ( kx ) + C
k x !1
2
1
k
= 1k sin !1 ( kx ) + C, x 2 <
2
dx
2
1
k
cos x + C
= sec !1 ( kx ) + C, k 2 x 2 > 1
F. Linearity Rules
1. Constant Multiple Rule:
2. Negative Rule:
! k f (x) dx = k ! f (x) dx, k
" ! f (x) dx = ! " f (x) dx, k
3. Sum or Difference Rule:
! ( f (x) ± g( x )) dx
=
! f (x) dx ± ! g(x) dx
G. Integrate the following:
! 12 dx =
1.
3.
! dx =
5.
! (9r
6.
![
7.
! (5
11
x
+ 1r )dr =
+ sin(3x)]dx =
)
t 3 +7 dt =
" 3
$# 4x 2 + 1 !
8.
(
9.
" (4e
10.
2.
4.
2
2!
is a constant
7
1! x
%
dx =
2 '
&
+ sec 2 ! )d! =
" sin! (cot ! + sec ! )d! =
2
! p dp =
3
! 2xdy =
11.
! (t
12.
"$ 4
7
1
5 %
' # x + x 3 ! 3x 4 + 1+ x 2 & dx =
3
)
2
+ 1 dt =
13.
! (e + 4
14.
! (x
15.
# 1 " sin !
16.
! tan
17.
% "r + r
t
3
t
- e2 + 5 )dt =
+ x ) dx
2
sin !
2
!
2
d!
y dy
1 #! 1 #
2 $ " r - 2 $ dr
r
IV.
Initia l Va lue Pro bl e ms
dy
= f (x) that has a derivative in it is called a diffe re nti al
dx
eq uat io n. The problem of finding a function y of x when we know its derivative and its
initial condition value y0 at a particular point x0 is called an init ial val ue pr o bl em .
A. De f n : An equation like
F(x) + C of the function f (x) gives the ge ne ra l
dy
so lut io n y = F(x) + C of the differential equation
= f (x) . The general solution gives
dx
all the solutions of the equation. When an initial value problem is solved, the p art ic ula r
so lut io n that satisfies the initial condition y(x0 ) = y0 is found.
B. The most general antiderivative
C. Examples
Find the initial function given the initial condition(s).
dy
= e x + 2x 4 ; y ( 0 ) = 7
1. dx
2.
f ! ( x ) = 2x "
3
; f
1 + x2
( 3) = 1
3. A ball is thrown upward from the top of a building 160 ft tall. If the ball’s upward
velocity at time t=1sec is 16 ft/sec, determine (a.) how long it takes the ball to reach
the ground below, (b.) the ball’s impact velocity. Ignore air resistance.
4. Given that the graph of f(x) passes through the pt
at the pt (x,f(x)) is sec x, find f.
2
( !4 ,3)
and the slope of its tangent
v (t ) = 4cos(t ) and that s(! ) = 1 , find the object’s position at time
t = !4 sec if distance is measured in meters.
5. Given that
6. Determine
f (x) given f !!(x) = 6x 2 " 10; f (0) = 9, f (2) = "35
f !(x) is given below, sketch the graph of f (x) if f (x) is continuous
and f (!3) = !6
7. The graph of
8.
A car braked with constant deceleration of 40
ft
s2
, producing skid marks measuring
160ft before coming to a stop. How fast was the car traveling when the brakes were
first applied?