Lesson 11

Lesson 11
Antiderivatives
Last time group work
dP
= 5P ! 50 and P(0) = 8
1- Find the solution for:
dt
5t
5t
P! = 5(P " 10), P(t) = 10 + Ce # P(t) = 10 " 2e
2- If an object takes 40 minutes to cool from 30 degrees to 24
degrees in a 20 degree room, how long will it take the object
to cool to 21 degrees?
Solve for t when y(t) = 21
y(0) = 20 + Cek!0 = 20 + C = 30 " C = 10
ln(2 / 5)
40k
y(40) = 24 = 20 + 10e
"k=
# -0.023
40
$ ln(0.1)
y(t) = 21 = 20 + 10exp(-0.023! t) " t =
# 100.5
0.023
Antiderivatives
Def: We say F is an antiderivative for f if F'(x) = f(x).
Example: x2 is an antiderivative for 2x, because d ( x 2 ) = 2x
dx
How about an antiderivative for x2?
First guess: Maybe x3…
d 3
x ) = 3x 2
(
dx
Check:
Oops… let’s put in an extra 1/3 to cancel this 3…
1
3
(
dx
x 3 is an antiderivative for x2, because d
1
3
x3) =
1
3
(3x ) = x
2
2
Antiderivatives
• Useful for solving integrals, through the
fundamental theorem.
• Also useful for solving some initial value
problems.
• We can always check the result by taking
the derivative, so let’s find a few by
guessing and checking…
Guess and Check
In your group, try to find antiderivatives for the nine functions
on p. 1 of your notes for Lesson 11. Make a good guess, then
check your result!
Antiderivatives
f(x) = x Antiderivative:
1
2
x2
Check:
f(x) = x5 Antiderivative:
1
6
6
Check:
f(x) = 3x7 Antiderivative:
3
8
f(x) = sin(x) Antiderivative:
x
x8
– cos(x)
d 1 2
(
2 x )=
dx
d 1 6
(
6 x )=
dx
Check: d
dx
(
3
8
x8) =
3
8
1
2
(2x ) = x
1
6
(6x ) = x
5
5
7
7
8x
=
3x
( )
Check: d (! cos( x ) ) = sin( x )
dx
d
f(x) = cos(x) Antiderivative: sin(x) Check:
(sin(x)) = cos(x)
dx
d
f(x) = ex Antiderivative: e x Check: (e x ) = e x
dx
d
1
f(x) = 1 Antiderivative: ln(x) Check:
ln(x)
=
(
)
dx
x
x
Antiderivatives
f(x) = 7 Antiderivative: 7x
Check:
d
(7x ) = 7
dx
Note: We only use x because that is the specified
variable. An antiderivative for f(t) = 7 would be
f(x) = x + ex
Antiderivative: 12 x 2 + e x (Just combine patterns we
already knew…)
Check:
d 1 2
x
x
+
e
=
(
)
2
dx
1
2
(2x ) + e x = x + e x
Antiderivatives Rules
We begin to see some patterns….
If f(x) is…
x
n
1
cos(x)
sin(x)
…then an antiderivative is…
x n +1 except if n = –1
x (assuming the variable is x!)
1
n +1
sin(x)
–cos(x)
ex
ex
1
x
ln x
(makes it work for
x < 0 too.)
Antiderivatives Rules
Antiderivatives are linear, just like derivatives.
So if F is an antiderivative for f and G is an antiderivative
for g, then an antiderivative for
b f(x) + c g(x)
is
b F(x) + c G(x)
Example
An antiderivative for
is
3(
1
1+1
x
1+1
3x + 7 ! 3e
) + 7x ! 3e
x
x
= x + 7x ! 3e
3
2
2
x
Example
To find an antiderivative for
3 3
+ 2 , rewrite as
x x
! 1$
3# & + 3x '2
" x%
Rewriting this would
be a bad idea, as our
power rule doesn’t
work for n = –1!
Then an antiderivative is easy:
3ln x + 3(
1
!1
3
x ) = 3ln x !
x
!1
Another Power Rule Example
x , rewrite as x1/ 2 .
To find an antiderivative for
Then an antiderivative is
1 12 +1
1 3/2 2 3/2
x =
x = 3x
1
3/2
2 +1
Is it correct? We can always check!
d 2 3/2
=
(
)
3 x
dx
2
3
(
3
2
x1/ 2 ) = x1/ 2 = x
The Indefinite Integral
We will use the notation
!
f (x) dx
to represent all possible antiderivatives of the function
f(x), with respect to the variable x.
Called the indefinite integral of f(x).
For example, what is
One antiderivative is
But so are
1
3
!
1
3
x 3 + 7 or
x 2 dx
x3
1
3
x 3 " ! We can add any constant…
The Indefinite Integral
In fact,
!
x 2 dx = 13 x 3 + C where C can be any constant.
To find the indefinite integral, find one antiderivative, then add “+ C”.
!
x n dx =
1
n +1
x n +1 + C (if n ≠ –1)
! k f (x) dx = k ! f (x) dx (constant multiple rule)
! f (x) + g(x) dx = ! f (x) dx + ! g(x) dx (sum rule)
Example
! 3sin( x ) + 2 dx =
! 2
"
q
( $ q + e # 6 % dq =
&
'
!3cos( x ) + 2 x + C
Using Algebra
How could we find the following:
" (x !1)(x + 1) dx
Rearrange the integrand algebraically first, then integrate:
2
1 3
(
x
!
1)(
x
+
1)
dx
=
x
!
1
dx
=
3 x ! x+C
"
"
This is a good general trick for products of polynomials.
! (3x + 1)
2
dx =
2
3
2
9 3
6 2
9
x
+
6
x
+
1
dx
=
x
+
x
+
x
+
C
=
3
x
+
3
x
+ x+C
3
2
!
Warning: We have no rule for integrals of products, so this is
pretty much all we can do with a product (for now).
Example: Quotients
x (1 + x )
x + x2
1 2
dx
=
1
+
x
dx
=
x
+
2 x +C
!
! x dx = ! x
1/ 2
t
t
1
(1/ 2) ! 2
!3/ 2
!3/ 2 +1
dt
=
dt
=
t
dt
=
t
dt
=
t
+C
! t2
" t2
"
"
!3/ 2 + 1
=
!1/ 2
1
!1/ 2
t
+ C = !2t !1/ 2 + C
Non-Example
Warning: Only multiplication cancels division. Don’t make
x
illegal “simplifications”!
dx
2
x +x
For now, we cannot find
!
***WRONG***
Can only cancel
common factors
x
1
1
=
! +1
2
x + x x +1 x
***WRONG***
1
1 1
! +
a+b a b
What About C ?
If we have a derivative, plus an initial condition, we can
solve for C.
Example
Find a function F(x) such that F'(x) = sin(x) + 3x,
and F(0) = 2.
First, find all possible functions for F:
3 2
sin(x)
+
3x
dx
=
"cos(x)
+
!
2 x +C
Now find C so that F(0) = 2:
F(0) = !cos(0) + 32 0 2 + C = !1+ C = 2
So C = 3, and F(x) = !cos(x) + 32 x 2 + 3
Example
Find a curve which has slope given by 3x at each value of x, and
which goes through the point (0, 2) in the plane.
3x 2
m = f !(x) = 3x
" f (x) = # 3x dx =
+C
2
3(0) 2
point : (0,2) $ f (0) = 2 =
+C
2
2=C
3x 2
" f (x) =
+2
2
InClass Assignment
Have a nice break!