Page 1 of 15 MCC@WCCUSD 05/03/2013
Connecting Algebra to Calculus Indefinite Integrals
Objective: Find Antiderivatives and use basic integral formulas to find Indefinite Integrals
and make connections to Algebra 1 and Algebra 2. Standards: Algebra 1 2.0, 10.0, 11.0/N-RN.1, N-RN.2, A-SSE.2, Calculus 15.0 Lesson: Suppose you were asked to find a function
€
F whose derivative is
€
f x( ) = 3x 2? Could you go back and find the original function whose derivative is
€
3x 2? Talk to a neighbor, and then I will ask for a non-volunteer to answer.
Since
€
ddx
x 3 = 3x 2, then F x( ) = x 3. We can check by taking the derivative of
€
F x( ) .
Check:
F x( ) = x3
!F x( ) = 3x2
Now, we can see that
!F x( ) = f x( )
= 3x2
We call
€
F an antiderivative of
€
f . “The function
€
F is an antiderivative of
€
f .” What if we tried saying that the original function is
€
F x( ) = x 3 + 5? Does this give us the correct derivative that we are looking for? What about
€
F x( ) = x 3 − 215? How many antiderivatives are there for
€
f x( ) = 3x 2? Well, since the derivative of a constant is equal to
€
0, there are infinitely many antiderivatives for the function
€
f x( ) = 3x 2. In general, we can write
€
F x( ) = x 3 + C since any constant C will result in
€
" F x( ) = f x( )= 3x 2
.
We can call
€
F x( ) = x 3 + C a “Family of Antiderivatives”. We can always check our work by taking the derivative
€
F .
Page 2 of 15 MCC@WCCUSD 05/03/2013
Exploration: Finding Antiderivatives With your partner or group, for each derivative, find the original function
€
F . In other words, find the antiderivative of
€
f . Justify your work. a)
€
f x( ) = 2x F x( ) = x2 +C!" #$
b)
€
f x( ) = x F x( ) = x2
2+C
!
"#
$
%&
c)
€
f x( ) = x 2 F x( ) = x3
3+C
!
"#
$
%&
d)
€
f x( ) =1x 2
F x( ) = − 1x+C
"
#$%
&'
e)
€
f x( ) =1x 3
F x( ) = − 12x2
+C"
#$%
&' Students can justify their work by showing that !F x( ) = f x( ) .
Notation for Indefinite Integrals:
f x( ) ∫ dx = F x( )+C
Definition of an Antiderivative A function
€
F is an antiderivative of f on an interval I if !F x( ) = f x( ) for all x in I. Note: An antiderivative is also called an Indefinite Integral.
Integral Sign Variable of Integration
Integrand Constant of Integration
Antiderivative of f
Page 3 of 15 MCC@WCCUSD 05/03/2013
We can substitute to show the “inverse” nature of differentiation and integration. Since,
€
" F x( ) = f x( ) and " F x( )∫ dx = F x( ) + C,
€
Then f x( )∫ dx = F x( ) + C
So, if f x( )∫ dx = F x( ) + C
Then ddx
f x( )∫ dx[ ] = f x( )
Some Basic Integration Rules (There is a student half-sheet handout at the end of the lesson.)
Differentiation Formula Integration Formula
€
ddx
C[ ] = 0
€
0 dx = C∫
€
ddx
kx[ ] = k
€
k dx = kx + C∫
Constant Multiple Rule for Derivatives
€
ddx
k ⋅ f x( )[ ] = k ⋅ # f x( )
Constant Multiple Rule for Integrals
€
k ⋅ f x( )∫ dx = k f x( )∫ dx
Sum and Difference Rule for Derivatives
€
ddx
f x( ) ± g x( )[ ] = " f x( ) ± " g x( )
Sum and Difference Rule for Integrals
€
f x( ) ± g x( )[ ]∫ dx = f x( ) dx ± g x( ) dx∫∫
Power Rule for Derivatives
€
ddx
xn[ ] = nxn−1
Power Rule for Integrals
xn dx = xn+1
n+1+C, n ≠ −1∫
Note: Notice that we do not have a Product Rule or Quotient Rule for Integrals at this time.
Page 4 of 15 MCC@WCCUSD 05/03/2013
Example: Find the indefinite integral using two different methods.
€
2x + 3x∫ dx
Method 1: Decomposition Method 2: Rewrite Quotient as a Product
€
2x + 3x∫ dx
=2xx
+3x
#
$ % &
' ( ∫ dx
=2x
x12
+3
x12
#
$
% %
&
'
( (
∫ dx
= 2x12 + 3x
−12
#
$ %
&
' ( ∫ dx
= 2 x12dx + 3 x
−12∫∫ dx
= 2 23x
32
#
$ %
&
' ( + 3 2
1x
12
#
$ %
&
' ( + C
=43x
32 + 6x
12 + C
=23x
12 2x + 9( ) + C
€
2x + 3x∫ dx
=2x + 3
x12
dx∫
= x−
12 2x + 3( )dx∫
= 2x12 + 3x
−12
$
% &
'
( ) dx∫
= 2 x12dx + 3 x
−12∫∫ dx
= 2 23x
32
*
+ ,
-
. / + 3 2
1x
12
*
+ ,
-
. / + C
=43x
32 + 6x
12 + C
=23x
12 2x + 9( ) + C
Check the result by differentiation:
ddx
23x
12 2x + 9( )+C
!
"#
$
%&
=ddx
43x
32 + 6x
12 +C
!
"#
$
%&
=32⋅43x
32−1+
12⋅6x
12−1+ 0
= 2x12 +3x
−12
= 2 x + 3x
=2 x
1⋅xx
)
*+
,
-.+
3x
=2xx+
3x
=2x +3
x which is our integrand!
Page 5 of 15 MCC@WCCUSD 05/03/2013
Class Group Activity In your group, find the indefinite integral using two methods. Check your result by differentiation. Your group will be asked to display your work. 1) 2) 3) 4) 5) 6)
€
x3 + 3x2∫ dx
€
x2 + 2x− 3x4 dx∫
€
x2 + x+1x∫ dx
€
x2 −1x3
dx∫
€
2x+12 x∫ dx
€
x4 + 5x2 − 7x3∫ dx
Page 6 of 15 MCC@WCCUSD 05/03/2013
Solutions to the Class Group Activity Problem 1 Method 1: Decomposition Method 2: Rewrite Quotient as a Product
€
x 3 + 3x 2∫ dx
=x 3
x 2 +3x 2
#
$ %
&
' ( dx∫
= x + 3x−2( )dx∫= xdx + 3x−2∫∫ dx
= xdx + 3 x−2∫∫ dx
=x 2
2+
3x−1
−1+ C
=x 2
2−
3x
+ C
=x 2
2⋅xx#
$ % &
' ( −
3x⋅
22#
$ % &
' ( + C
=x 3
2x−
62x
+ C
=x 3 − 6
2x+ C
€
x 3 + 3x 2∫ dx
= x−2 x 3 + 3( )dx∫= x + 3x−2( )dx∫= xdx + 3x−2∫∫ dx
= xdx + 3 x−2∫∫ dx
=x 2
2+
3x−1
−1+ C
=x 2
2−
3x
+ C
=x 2
2⋅xx%
& ' (
) * −
3x⋅
22%
& ' (
) * + C
=x 3
2x−
62x
+ C
=x 3 − 6
2x+ C
Check the result by differentiation:
ddx
x3 − 62x
+C"
#$
%
&'
=ddx
x3
2x−
62x
+C"
#$
%
&'
=ddx
12x2 −3x−1 +C
"
#$%
&'
=21⋅12x2−1 −3⋅−1x−1−1 + 0
= x +3x−2
= x + 3x2
=x1⋅x2
x2
)
*+
,
-.+
3x2
=x3 +3x2 which is our integrand!
Page 7 of 15 MCC@WCCUSD 05/03/2013
Problem 2 Method 1: Decomposition Method 2: Rewrite Quotient as a Product
2
4
2
4 4 4
2 4 1 4 4
2 3 4
2 3 4
1 2 3
2 3
2 3
2 3
2 3
2 3
2 3
2 31 2 31 1 1
x x dxx
x x dxx x x
x x x dx
x x x dx
x dx x dx x dx
x x x C
Cx x x
− − −
− − −
− − −
− − −
+ −
⎡ ⎤= + −⎢ ⎥
⎣ ⎦⎡ ⎤= + −⎣ ⎦⎡ ⎤= + −⎣ ⎦
= + −
⎡ ⎤ ⎡ ⎤= + − +⎢ ⎥ ⎢ ⎥− − −⎣ ⎦ ⎣ ⎦
= − − + +
∫
∫
∫∫∫ ∫ ∫
( )( )
2
4
4 2
2 3 4
2 3 4
2 1 3 1 4 1
2 31
1 2 3
2 3
2 3
2 3
2 3
2 3
2 32 1 3 1 4 1
2 32 3
1 1 1
x x dxx
x x x dx
x x x dx
x dx x dx x dx
x x x C
x xx C
x x x C
Cx x x
−
− − −
− − −
− + − + − +
− −−
− − −
+ −
= + −
= + −
= + −
⎡ ⎤ ⎡ ⎤= + − +⎢ ⎥ ⎢ ⎥− + − + − +⎣ ⎦ ⎣ ⎦
⎡ ⎤ ⎡ ⎤= − + − +⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦= − − + +
= − − + +
∫∫∫∫ ∫ ∫
Check the result by differentiation:
( ) ( ) ( )
( )
2 3
1 2 3
1 1 2 1 3 1
2 3 4
4 2
2
4
1 1 1
1 2 3 0
2 3
2 3
2 3 which is our integrand!
d Cdx x x xd x x x Cdx
x x x
x x x
x x x
x xx
− − −
− − − − − −
− − −
−
⎡ ⎤− − + +⎢ ⎥⎣ ⎦
⎡ ⎤= − − + +⎣ ⎦
= − − − − + − +
= + −
= + −
+ −=
Page 8 of 15 MCC@WCCUSD 05/03/2013
Problem 3: Method 1: Decomposition Method 2: Rewrite Quotient as a Product
x2 + x +1x∫ dx
= x2
x+ x
x+ 1
x
⎡
⎣⎢
⎤
⎦⎥∫ dx
= x2
x12
+ x
x12
+ 1
x12
⎡
⎣⎢⎢
⎤
⎦⎥⎥∫ dx
= x32 + x
12 + x
− 12
⎡
⎣⎢
⎤
⎦⎥∫ dx
= x32∫ dx + x
12∫ dx + x
− 12∫ dx
= 25
x52 + 2
3x
32 + 2
1x
12 +C
= 215
x12 3x2 +5x +15( ) +C
x2 + x +1x∫ dx
= x2 + x +1
x12
∫ dx
= x− 1
2 x2 + x +1⎡⎣ ⎤⎦∫ dx
= x32 + x
12 + x
− 12
⎡
⎣⎢
⎤
⎦⎥∫ dx
= x32∫ dx + x
12∫ dx + x
− 12∫ dx
= 25
x52 + 2
3x
32 + 2
1x
12 +C
= 215
x12 3x2 +5x +15( ) +C
Check the result by differentiation:
ddx
215
x12 3x2 +5x +15( ) +C
⎡
⎣⎢
⎤
⎦⎥
= ddx
25
x52 + 2
3x
32 + 2x
12 +C
⎡
⎣⎢
⎤
⎦⎥
= 52⋅ 25
x52−1+ 3
2⋅ 23
x32−1+ 1
2⋅2x
12−1+ 0
= x32 + x
12 + x
− 12
= xx12 x
12
x12
⎛
⎝
⎜⎜
⎞
⎠
⎟⎟+ x
1i
x
x
⎛
⎝⎜
⎞
⎠⎟ +
1
x
= x2
x+ x
x+ 1
x
= x2 + x +1
xwhich is our integrand!
Page 9 of 15 MCC@WCCUSD 05/03/2013
Problem 4:
Method 1: Decomposition Method 2: Rewrite Quotient as a Product
( )( )
2
3
2
3 3
2
3 32 2
1 32 2
1 32 2
1 2 3 22 2 2 2
3 12 2
122
2
1
1
1
2 23 12 232 332 3
3
x dxxx dxx x
x dxx x
x x dx
x dx x dx
x x C
x x C
x x C
xC
x
−
−
+ − +
−
−
−
⎡ ⎤= −⎢ ⎥
⎣ ⎦⎡ ⎤⎢ ⎥= −⎢ ⎥⎣ ⎦⎡ ⎤
= −⎢ ⎥⎣ ⎦
= −
⎛ ⎞= − − +⎜ ⎟⎝ ⎠
= + +
= + +
+= +
∫
∫
∫
∫
∫ ∫
x2 −1
x3∫ dx
=x2 −1
x32
∫ dx
= x−32 x2 −1( ) dx∫
= x−32+42 − x
−32 1( )
#
$%%
&
'((
∫ dx
= x12 dx − x
−32 dx∫∫
=23x12+22 − −
21
)
*+
,
-.x
−32+22 +C
=23x32 + 2x
−12 +C
=23x−12 x2 +3( )+C
=2 x2 +3( )3 x
+C
Checking the result by differentiation is shown on the next page.
Page 10 of 15 MCC@WCCUSD 05/03/2013
Check the result by differentiation:
ddx
2 x2 +3( )3 x
+C!
"
##
$
%
&&
=ddx
23x−
12 x2 +3( )+C
!
"##
$
%&&
=ddx
23x
32 + 2x
−12 +C
!
"##
$
%&&
=32
23x
32−
22
(
)**
+
,--+ −
12
(
)*
+
,- 2x
−12−
22
(
)**
+
,--+0
= x12 − x
−32
= x12 x
12
x12
(
)
***
+
,
---−
1
x32
=x
x12
−1
x32
=x
x12
x22
x22
(
)
***
+
,
---−
1
x32
=x2
x32−
1
x32
=x2 −1
x32 which is our integrand!
Page 11 of 15 MCC@WCCUSD 05/03/2013
Problem 5: Method 1: Decomposition Method 2: Rewrite Quotient as a Product
2x +12 x
∫ dx
=2x2 x
+12 x
"
#$
%
&'∫ dx
=2x
2x12
+1
2x12
"
#
$$
%
&
''∫ dx
= x12 +12x−12
"
#$$
%
&''
∫ dx
= x12 dx +∫ 1
2x−12 dx∫
=23x12+22 +1221x−12+22
)
*++
,
-..+C
=23x32 + x
12 +C
=13x12 2x +3( )+C
( )
( )
12
12
11 22
1 12 2
1 2 1 22 2 2 2
3 12 2
12
2 122 1
2
2 12
2
12
2 1 23 2 1
231 2 33
x dxx
x dxx
x x dx
xx dx
x dx x dx
x x C
x x C
x x C
−
−
−
+ − +
+
+=
= +
⎛ ⎞⎜ ⎟= +⎜ ⎟⎜ ⎟⎝ ⎠
= +
⎛ ⎞= + +⎜ ⎟
⎝ ⎠
= + +
= + +
∫
∫
∫
∫
∫ ∫
Check the result by differentiation: ddx
13x
12 2x +3( )+C
!
"##
$
%&&
=ddx
23x
32 + x
12 +C
!
"##
$
%&&
=32
23x
32−
22
(
)**
+
,--+
12x
12−
22
(
)**
+
,--+0
= x12 +
12x−
12
= x12 +
1
2x12
=x
1x
x
(
)**
+
,--+
1
2 x
=x
x
22
(
)*+
,-+
1
2 x
=2x +1
2 x which is our integrand!
Page 12 of 15 MCC@WCCUSD 05/03/2013
Problem 6:
Method 1: Decomposition Method 2: Rewrite Quotient as a Product
x4 +5x2 −7x3
∫ dx
=x4
x13
+5x2
x13
−7
x13
#
$
%%
&
'
((∫ dx
= x113 +5x
53 −7x
−13
#
$%%
&
'((
∫ dx
= x113 dx +5 x
53 dx∫ −7 x
−13 dx∫∫
=314x113+33 +5 3
8x53+33
#
$%%
&
'((−7 3
2x−13+33
#
$%%
&
'((+C
=314x143 +158x83 −212x23 +C
=356x23 4x4 +35x2 −196( )+C
x4 +5x2 −7x3∫ dx
= x−
13 x4 +5x2 −7( )∫ dx
= x113 +5x
53 −7x
−13
#
$%%
&
'((∫ dx
= x113∫ dx +5 x
53∫ dx −7 x
−13 dx∫
=3
14x
113+
33 + 5⋅ 3
8#
$%
&
'(x
53+
33 − 7 ⋅ 3
2#
$%
&
'(x
−13+
33 +C
=3
14x
143 +
158x
83 −
212x
23 +C
=3
56x
23 4x4 +35x2 −196( )+C
Check the result by differentiation: ddx
356
x23 4x4 +35x2 −196( )+C
"
#$
%
&'
=ddx
314
x143 +
158x
83 −
212x
23 +C
"
#$
%
&'
= x113 +
83⋅158
)
*+
,
-.x
53 −
23⋅212
)
*+
,
-.x
−13 + 0
= x113 + 5x
53 − 7x
−13
=x
113
1+
5x53
1−
7
x13
=x
113
1⋅x
13
x13
)
*
+++
,
-
.
..+
5x53
1⋅x
13
x13
)
*
+++
,
-
.
..−
7
x13
=x4 + 5x2 − 7
x13
=x4 + 5x2 − 7
x3 which is our integrand!
Page 13 of 15 MCC@WCCUSD 05/03/2013
Differentiation Formula Integration Formula
€
ddx
C[ ] = 0
€
0 dx = C∫
€
ddx
kx[ ] = k
€
k dx = kx + C∫
Constant Multiple Rule for Derivatives
€
ddx
k ⋅ f x( )[ ] = k ⋅ # f x( )
Constant Multiple Rule for Integrals
€
k ⋅ f x( )∫ dx = k f x( )∫ dx
Sum and Difference Rule for Derivatives
€
ddx
f x( ) ± g x( )[ ] = " f x( ) ± " g x( )
Sum and Difference Rule for Integrals
€
f x( ) ± g x( )[ ]∫ dx = f x( ) dx ± g x( ) dx∫∫
Power Rule for Derivatives
€
ddx
xn[ ] = nxn−1
Power Rule for Integrals
xn dx = xn+1
n+1+C, n ≠ −1∫
Differentiation Formula Integration Formula
€
ddx
C[ ] = 0
€
0 dx = C∫
€
ddx
kx[ ] = k
€
k dx = kx + C∫
Constant Multiple Rule for Derivatives
€
ddx
k ⋅ f x( )[ ] = k ⋅ # f x( )
Constant Multiple Rule for Integrals
€
k ⋅ f x( )∫ dx = k f x( )∫ dx
Sum and Difference Rule for Derivatives
€
ddx
f x( ) ± g x( )[ ] = " f x( ) ± " g x( )
Sum and Difference Rule for Integrals
€
f x( ) ± g x( )[ ]∫ dx = f x( ) dx ± g x( ) dx∫∫
Power Rule for Derivatives
€
ddx
xn[ ] = nxn−1
Power Rule for Integrals
xn dx = xn+1
n+1+C, n ≠ −1∫
Page 14 of 15 MCC@WCCUSD 05/03/2013
Warm Up
CCSS: Calculus 4.4 Review: Calculus 4.0
Which of the following are true for this
polynomial Recall: Power Rule for Derivatives:
Power Rule for Derivatives: Given:
1n nd x nxdx
−=
Find the derivative:
a) 5d x
dx
b) d xdx
c) 4
1ddx x
⎡ ⎤⎢ ⎥⎣ ⎦
( ) 3 22 6 18 1?f x x x x= − + + +
( )( ) 1
n
n
f x x
f x nx −
=
′ =
A. ( ) 3 26 12 18f x x x x′ = − + + True
False
B. ( ) 26 12 18f x x x′ = − + + True
False
C. ( ) 22 3 183
f x x x′ = − − + True
False
D. !f x( ) = −6 x +1( ) x −3( ) True
False
E. ( ) ( )( )6 1 2 3f x x x′ = − − + True
False
x
y
Page 15 of 15 MCC@WCCUSD 05/03/2013
Solutions to Warm-Up Quadrant I a) b) c) Quadrant II f x( ) = −2x3 +6x2 +18x +1
!f x( ) = −6x2 +12x +18 → This makes choice B true
"f x( ) = −6 x2 − 2x −3( )!f x( ) = −6 x +1( ) x −3( ) → This makes choice D true
Choices A, C and E are all false
ddxx5 = 5x4
ddx
x
=ddxx12
=12x12−1
=12x−12
=12 x
ddx
1x4
=ddxx−4
= −4x−4−1
= −4x−5
= −4x5
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