section 6-1 antiderivatives and indefinite integralskim/math144_ch6_lecture.pdf · section 6-1...
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Calculus Chapter 6
6-1
Name ________________________________ Date ______________ Class ____________
Goal: To find antiderivatives and indefinite integrals of functions using the formulas and properties
In Problems 1–3, find each indefinite integral and check by differentiating.
1. 6x dx∫
2
2
6
6
2
3
y x dx
xy C
y x C
= ∫
= +
= +
Check: 23
6 0
6
y x C
dyx
dxdy
xdx
= +
= +
=
Section 6-1 Antiderivatives and Indefinite Integrals
Theorem 1 Antiderivatives If the derivative’s of two functions are equal on an open interval (a, b), then the functions differ by at most a constant. Symbolically, if F and G are differentiable functions on the interval (a, b) and '( ) '( )F x G x= for all x in (a, b), then ( ) ( )F x G x k= + for some constant k. Formulas and Properties of Indefinite Integrals For C and k both a constant
1. 1
,1
nn x
x dx Cn
+= +∫ +
1n ≠ −
2. x xe dx e C= +∫
3. 1
ln ,dx x Cx
= +∫ 0x ≠
4. ( ) ( )kf x dx k f x dx=∫ ∫
5. [ ( ) ( )] ( ) ( )f x g x dx f x dx g x dx± = ±∫ ∫ ∫
Calculus Chapter 6
6-2
2. 1
29x dx∫
12
32
32
32
32
9
9
2 9
3 1
6
y x dx
xy C
xy C
y x C
= ∫
= +
= ⋅ +
= +
Check: 3
2
12
12
6
3 60
2 1
9
y x C
dy x
dxdy
xdx
= +
= ⋅ +
=
3. 7 xe dx∫
7
7
x
x
y e dx
y e C
= ∫
= +
Check: 7
7 0
7
x
x
x
y e C
dye
dxdy
edx
= +
= +
=
4. Find all the antiderivatives for 17 4.dy
zdz
−= +
1
1
7
7
7 4
(7 4)
( 4)
( 4)
7 ln 4
z
z
dyz
dz
dy z dz
dy dz
dy dz
y z z C
−
−
= +
= +
= +
= +∫ ∫
= + +
Calculus Chapter 6
6-3
In Problems 5–8, find each indefinite integral.
5. 3 2( 2 8)x x x dx+ −∫
3 2 5 4 3
6 5 4
6 54
( 2 8) ( 2 8 )
2 8
6 5 4
22
6 5
x x x dx x x x dx
x x xC
x xx C
+ − = + −∫ ∫
= + − +
= + − +
6. 34
8x dx
x
⎛ ⎞−∫ ⎜ ⎟⎝ ⎠
( )32
52
3 44
3
52
5
3
88
8
3
8 2
53
x dx x x dxx
x xC
xC
x
−
−
⎛ ⎞− = −∫ ∫⎜ ⎟⎝ ⎠
= − +−
= − − +
7. 4
8 5xdx
x
+⎛ ⎞∫ ⎜ ⎟⎝ ⎠
( )4 3 4
3 4
2 3
2 3
8 5 8 5
8 5
8 5
2 34 5
3
xdx dx
x x x
x x dx
x xC
Cx x
− −
− −
+⎛ ⎞ ⎛ ⎞= +∫ ∫⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= +∫
= + +− −
= − − +
Calculus Chapter 6
6-4
8. 5 6
6
5 3 xx x edx
x
⎛ ⎞+∫ ⎜ ⎟⎝ ⎠
5 6
6
5 3 53
5ln 3
xx
x
x x edx e dx
xx
x e C
⎛ ⎞+ ⎛ ⎞= +∫ ∫ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠
= + +
In Problems 9–12, find the particular antiderivative of each derivative that satisfies the given conditions.
9. 3 2'( ) 4 6 3;R x x x= + + (1) 12R =
First find the indefinite integral of the function.
3 2
4 3 1
4 3
( ) (4 6 3)
4 6 3( )
4 3 1
( ) 2 3
R x x x dx
x x xR x C
R x x x x C
= + +∫
= + + +
= + + +
Using the condition given, find the specific value of C.
4 3
4 3
( ) 2 3
12 (1) 2(1) 3(1)
12 1 2 3
12 6
6
R x x x x C
C
C
C
C
= + + +
= + + += + + += +=
Substituting the specific value of C yields the particular equation
4 3( ) 2 3 6.R x x x x= + + +
Calculus Chapter 6
6-5
10. 3 3 5;tdye t
dt= + − (0) 5y =
First find the indefinite integral of the function.
2 1
2
(3 3 5)
(3 3 5)
3 53
2 1
33 5
2
t
t
t
t
dy e t dt
dy e t dt
t ty e C
ty e t C
= + −
= + −∫ ∫
= + − +
= + − +
Using the condition given, find the specific value of C.
2
20
33 5
2
3(0)5 3 5(0)
25 3 0 0
5 3
2
t ty e t C
e C
C
C
C
= + − +
= + − +
= + + += +=
Substituting the specific value of C yields the particular equation
233 5 2.
2t t
y e t= + − +
Calculus Chapter 6
6-6
11. 2
2
5 9;
dD x
dx x
−= (9) 50D =
First find the indefinite integral of the function.
2
2
2
2
1 1
5 9
(5 9 )
(5 9 )
5 9( )
1 19
( ) 5
xdD dx
x
dD x dx
dD x dx
x xD x C
D x x Cx
−
−
−
⎛ ⎞−= ⎜ ⎟⎝ ⎠
= −
= −∫ ∫
= − +−
= + +
Using the condition given, find the specific value of C.
9
( ) 5
950 5(9)
950 45 1
50 46
4
D x x Cx
C
C
C
C
= + +
= + +
= + += +=
Substituting the specific value of C yields the particular equation 9
( ) 5 4.D x xx
= + +
Calculus Chapter 6
6-7
12. 1 2'( ) 6 7 ;h x x x− −= + (1) 3h =
First find the indefinite integral of the function.
1 2
26
1
'( ) 6 7
( ) ( 7 )
7( ) 6 ln
17
( ) 6 ln
x
h x x x
h x x dx
xh x x C
h x x Cx
− −
−
−
= +
= +∫
= − +−
= + +
Using the condition given, find the specific value of C.
7
( ) 6 ln
73 6ln 1
13 6(0) 7
3 7
4
h x x Cx
C
C
C
C
= + +
= + +
= + += +
− =
Substituting the specific value of C yields the particular equation
7( ) 6 ln 4.h x x
x= + −
Calculus Chapter 6
6-8
Calculus Chapter 6
6-9
Name ________________________________ Date ______________ Class ____________
Goal: To find the indefinite integrals using general indefinite integral formulas
Section 6-2 Integration by Substitution
Formulas: General Indefinite Integral Formulas
1. 1[ ( )]
[ ( )] '( ) ,1
nn f x
f x f x dx Cn
+= +∫ +
1n ≠ −
2. ( ) ( )'( )f x f xe f x dx e C= +∫
3. 1
'( ) ln ( )( )
f x dx f x Cf x
= +∫
4. 1
,1
nn u
u du Cn
+= +∫ +
1n ≠ −
5. u ue du e C= +∫
6. 1
lndu u Cu
= +∫
Definition: Differentials If ( )y f x= defines a differentiable function, then
1. The differential dx of the independent variable x is an arbitrary real number. 2. The differential dy of the dependent variable y is defined as the product of
'( )f x and dx: '( )dy f x dx= Procedure: Integration by Substitution
1. Select a substitution that appears to simplify the integrand. In particular, try to select u so the du is a factor in the integrand.
2. Express the integrand entirely in terms of u and du, completely eliminating the original variable and its differential.
3. Evaluate the new integral if possible. 4. Express the antiderivative found in step 3 in terms of the original variable.
Calculus Chapter 6
6-10
In Problems 1–8, find each indefinite integral and check the result by differentiating.
1. 2 3(6 3 5) (12 3)x x x dx+ − +∫
Let 26 3 5,u x x= + − therefore (12 3) .du x dx= + Rewrite the original integral in terms of the variable u and solve.
2 3
3
4
2 4
(6 3 5) (12 3)
41
(6 3 5)4
y x x x dx
y u du
uy C
y x x C
= + − +∫
= ∫
= +
= + − +
Check:
2 4
2 3 2
2 3
1(6 3 5)
41
(4)(6 3 5) (6 3 5)4
(6 3 5) (12 3)
y x x C
dy dx x x x
dx dxdy
x x xdx
= + − +
= + − + −
= + − +
2. 3 23 ( 3 5)(3 3)x x x dx+ − +∫
Let 3 3 5,u x x= + − therefore 2(3 3) .du x dx= + Rewrite the original integral in terms of the variable u and solve.
13
13
43
43
3 23
3 2
43
3
3 43
( 3 5)(3 3)
( 3 5) (3 3)
3( 3 5)
43
( 3 5)4
y x x x dx
y x x x dx
y u du
uy C
y x x C
y x x C
= + − +∫
= + − +∫
= ∫
= +
= + − +
= + − +
Calculus Chapter 6
6-11
Check:
43
13
13
3 43
3
3 3
2 2
3 23
3( 3 5)
43
( 3 5)43 4
( 3 5) ( 3 5)4 3
(6 3 5) (3 3)
( 3 5)(3 3)
y x x C
y x x C
dy dx x x x
dx dx
dyx x x
dxdy
x x xdx
= + − +
= + − +
⎛ ⎞= + − + −⎜ ⎟⎝ ⎠
= + − +
= + − +
3. 3
4
2
2 16 1
tdt
t t
+∫
+ −
Let 42 16 1,u t t= + − therefore 3 3(8 16) 8( 2) .du t dt du t dt= + ⇒ = + Rewrite the original integral in terms of the variable u and solve.
3
4
4
1 8( 2)
8 2 16 11 1
81
ln8
ln 2 16 1
8
ty dt
t t
y duu
y u C
t ty C
+= ∫+ −
= ∫
= +
+ −= +
Check:
4
44
34
3
4
ln 2 16 1
81
(2 16 1)8(2 16 1)
1(8 16)
8(2 16 1)
2
2 16 1
t ty C
dy dt t
dt dxt t
dyt
dt t t
dy t
dt t t
+ −= +
= + −+ −
= ++ −
+=+ −
Calculus Chapter 6
6-12
4. 0.1xe dx−∫ Let 0.1 ,u x= − therefore 0.1 .du dx= − Rewrite the original integral in terms of the variable u and solve.
0.1
0.1
0.1
1( 0.1 )
0.11
0.11
0.1
10
x
x
u
u
x
y e dx
y e dx
y e du
y e C
y e C
−
−
−
= ∫
= − −∫
= − ∫
= − +
= − +
Check:
0.1
0.1
0.1
0.1
10
10 ( 0.1 )
10 ( 0.1)
x
x
x
x
y e C
dy de x
dt dxdy
edtdy
edt
−
−
−
−
= − +
= − −
= − −
=
5. 7( 7)x x dx+∫ Let 7,u x= + therefore .du dx= If we rewrite the original integral in terms of the variable u, the substitution would not be complete. We also need 7 .u x− = Now rewrite the original integral in terms of u and solve.
7
7
8 7
9 8
9 8
( 7)
( 7)
( 7 )
7
9 8
( 7) 7( 7)
9 8
y x x dx
y u u du
y u u du
u uy C
x xy C
= +∫
= −∫
= −∫
= − +
+ += − +
Calculus Chapter 6
6-13
Check:
9 8
8 7
8 7
7
7
( 7) 7( 7)
9 81 7
(9)( 7) ( 7) (8)( 7) ( 7)9 8
( 7) 7( 7)
( 7) ( 7 7)
( 7)
x xy C
dy d dx x x x
dt dx dxdy
x xdtdy
x xdtdy
x xdt
+ += − +
= + + − + +
= + − +
= + + −
= +
6. 2 42(ln(3 ))x
dxx
∫
Let 2ln(3 ),u x= therefore 2
6 2.
3
xdu dx du dx
xx= ⇒ = Rewrite the original integral in
terms of the variable u and solve.
( )
2 4
4
5
52
2(ln(3 ))
5
ln 3
5
xy dx
x
y u du
uy C
xy C
= ∫
= ∫
= +
= +
Check:
( )52
2 4 2
2 42
2 4
ln 3
51
(5)(ln 3 ) (ln 3 )5
6(ln 3 )
3
2(ln 3 )
xy C
dy dx x
dt dxdy x
xdt x
xdy
dt x
= +
=
=
=
Calculus Chapter 6
6-14
7. 1
2
31
xe dxx
−∫
Let 2,u x−= − therefore 33 22 .
xdu x dx du dx−= ⇒ = Rewrite the original integral in
terms of the variable u and solve.
1
2
12
3
3
1
1 2
2
1
2
x
x
u
y e dxx
y e dxx
y e du
−
−
= ∫
= ⋅∫
= ∫
12
1
21
2x
uy e C
y e C−
= +
= +
Check:
1
2
12
12
12
2
3
3
1
21
( )21
(2 )2
x
x
x
x
y e C
dy de x
dt dx
dye x
dt
dy e
dt x
−
− −
− −
−
= +
= −
=
=
Calculus Chapter 6
6-15
8. 2 3 5 2 3 512 (2 7) 12 (2 7)dy
x x dy x x dxdx
= + ⇒ = +
Let 32 7,u x= + therefore 26 .du x dx= Rewrite the original integral in terms of the variable u and solve.
2 3 5
3 5 2
5
6
3 6
12 (2 7)
2(2 7) 6
2
2
61
(2 7)3
y x x dx
y x x dx
y u du
uy C
y x C
= +∫
= +∫
= ∫
= +
= + +
Check:
3 6
3 5 3
3 5 2
2 3 5
1(2 7)
31
(6)(2 7) (2 7)3
2(2 7) (6 )
12 (2 7)
y x C
dy dx x
dt dxdy
x xdtdy
x xdt
= + +
= + +
= +
= +
9. The indefinite integral can be found in more than one way. Given the integral, 2 22 ( 3) ,x x dx+∫ first use the substitution method to find the indefinite integral and
then find it without using substitution.
Using substitution, let 2 3,u x= + therefore 2 .du x dx= Rewrite the original integral in terms of the variable u and solve.
2 2
2
3
2 3
2 ( 3)
31
( 3)3
y x x dx
y u du
uy C
y x C
= +∫
= ∫
= +
= + +
Calculus Chapter 6
6-16
Expanding the final function yields the following:
2 3
2 2 2
6 4 2
64 2
1( 3)
31
( 3)( 3)( 3)31
( 9 27 27)3
3 93
y x C
y x x x C
y x x x C
xy x x C
= + +
= + + + +
= + + + +
= + + +
Note: The “new” value of C above is the addition of the constant 9 and the original constant C from the integration. Without using substitution to find the given indefinite integral, you first would multiply the integrand as follows
2 2 2 2
4 2
5 3
2 ( 3) 2 ( 3)( 3)
2 ( 6 9)
2 12 18
x x x x x
x x x
x x x
+ = + +
= + +
= + +
Now find the indefinite integral using the above function as the integrand.
5 3
6 4 2
64 2
(2 12 18 )
2 12 18
6 4 2
3 93
y x x x dx
x x xy C
xy x x C
= + +∫
= + + +
= + + +
Calculus Chapter 6
6-17
Name ________________________________ Date ______________ Class ____________
Goal: To solve differential equations that involve growth and decay.
Section 6-3 Differential Equations; Growth and Decay
Theorem 1: Exponential Growth Law
If dQ
rQdt
= and 0(0) ,Q Q= then 0 ,rtQ Q e=
where 0Q = amount of Q at 0t =
r = relative growth rate (expressed as a decimal) t = time Q = quantity at time t Table 1: Exponential Growth
Description Model Solution Unlimited growth dy
dtky=
, 0k t > (0)y c=
kty ce=
Exponential decay dydt
ky= −
, 0k t > (0)y c=
kty ce−=
Limited growth ( )dydt
k M y= −
, 0k t > (0) 0y =
(1 )kty M e−= −
Logistic Growth ( )dydt
ky M y= −
, 0k t >
1(0) Mc
y +=
1 kMt
My
ce−=
+
Calculus Chapter 6
6-18
In Problems 1–4, find the general or particular solution, as indicated, for each differential equation.
1. 36dy
xdx
−=
The differential equation can be found by using the integration properties from Section 6.1.
3
2
2
6
6
2
3
y x dx
xy C
y x C
−
−
−
= ∫
= +−
= − +
2. 323 ;xdy
x edx
−= − (0) 4y =
The differential equation can be found by using the integration properties from
Section 6.2. Let 3,u x= − therefore 23 .du x dx= − Rewrite the original integral in terms of the variable u and solve.
3
3
23 x
u
u
x
y x e dx
y e du
y e C
y e C
−
−
= −∫
= ∫
= +
= +
Using the condition given, find the specific value of C.
3
3(0)
0
4
4
4 1
3
xy e C
e C
e C
C
C
−
−
= +
= +
= += +=
Substituting the specific value of C yields the particular equation 3
3.xy e−= +
Calculus Chapter 6
6-19
3. 6dy
ydx
= −
The differential equation is in the form of exponential decay; therefore, using the exponential decay model in Table 1 yield’s the following solution:
6xy ce−=
4. 4 ;dx
xdt
= (0) 2x =
The differential equation is in the form of unlimited growth; therefore, using the unlimited growth model in Table 1 yield’s the following solution:
4tx ce= Using the condition given, find the specific value of C.
4
4(0)
0
2
2
2
tx ce
ce
ce
c
=
=
==
Substituting the specific value of C yields the particular solution 42 .tx e=
5. Find the amount A in an account after t years if
0.07dA
Adt
= and (0) 8000A =
The exponential growth law (unlimited growth) applies to the situation. Since we are given the initial amount, the amount in an account after t years would be:
0.078000 tA e=
Calculus Chapter 6
6-20
6. A single injection o a drug is administered to a patient. The amount Q in the body decreases at a rate proportional to the amount present. For a particular drug, the rate
is 6% per hour. Thus, 0.06dQ
Qdt
= − and 0(0)Q Q= where t is time in hours.
a. If the initial injection is 5 milliliters [ (0) 5Q = ], find ( )Q Q t= satisfying both conditions.
b. How many milliliters (to two decimal places) are in the body after 8 hours? c. How many hours (to two decimal places) will it take for half the drug to be left in
the body?
a. By the exponential growth law (exponential decay) for the given conditions, we
have the equation 0.06( ) 5 .tQ t e−= b. After 8 hours, 8,t = therefore
0.06(8)
0.48
(8) 5
(8) 5
(8) 5(0.61878)
(8) 3.09
Q e
Q e
Q
Q
−
−
=
==≈
Therefore, after 8 hours, approximately 3.09 milliliters will remain in the body.
c. Half the initial injection is 2.5 milliliters. We need to find t such that ( ) 2.5.Q t =
0.06
0.06
0.06
0.06
( ) 5
2.5 5
0.5
ln 0.5 ln
ln 0.5 0.06
11.55
t
t
t
t
Q t e
e
e
e
t
t
−
−
−
−
=
=
=
== −≈
Therefore, it will take approximately 11.55 hours for the amount of the drug to be half of the initial amount.
Calculus Chapter 6
6-21
7. A company is trying to expose a new product to as many people as possible through radio ads. Suppose that the rate of exposure to new people is proportional to the number of those who have not heard of the product out of L possible listeners. No one is aware of the product at the start of the campaign, and after 15 days, 50% of L
are aware of the product. Mathematically, ( ),dN
k L Ndt
= − (0) 0,N = and
(15) 0.5 .N L= a. Solve the differential equation. b. What percent of L will have been exposed after 7 days of the campaign? c. How many days will it take to expose 75% of L?
a. By the exponential growth law (limited growth) for the given conditions, we have
the equation
15
15
15
15
15
( ) (1 )
0.5 (1 )
0.5 1
0.5
0.5
ln 0.5 ln
ln 0.5 15
0.046
kt
k
k
k
k
k
N t L e
L L e
e
e
e
e
k
k
−
−
−
−
−
−
= −
= −
= −
− = −
=
== −=
Therefore, the solution is 0.046( ) (1 ).tN t L e−= −
b. After 7 days, 7,t = therefore
0.046(7)(7) (1 )
(7) (1 0.724698)
(7) (0.275302)
(7) 0.275
N L e
N L
N L
N L
−= −= −=≈
Therefore, after 7 days, approximately 27.5% of L will have been exposed to the product.
Calculus Chapter 6
6-22
c. Find t such that ( ) 0.75 .N t L=
0.046
0.046
0.046
0.046
0.046
0.046
( ) (1 )
0.75 (1 )
0.75 1
0.25
0.25
ln 0.25 ln
ln 0.25 0.046
30.1
t
t
t
t
t
t
N t L e
L L e
e
e
e
e
t
t
−
−
−
−
−
−
= −
= −
= −
− = −
=
== −≈
Therefore, it will take approximately 30 days to expose 75% of L to the new product.
Calculus Chapter 6
6-23
Name ________________________________ Date ______________ Class ____________
Goal: To calculate the values of definite integrals using the properties.
Section 6-4 The Definite Integral
Theorem: Limits of Left and Right Sums If ( ) 0f x > and is either increasing or decreasing on [a, b], then its left and right sums approach the same real number as .n →∞ Theorem: Limit of Riemann Sums If f is a continuous function on [a, b], then the Riemann sums for f on [a, b] approach a real number limit I as .n →∞ Definition: Definite Integral Let f be a continuous function on [a, b]. The limit I of Riemann sums for f on [a, b] is
called the definite integral of f from a to b and is denoted as ( ) .ba f x dx∫
Properties: Definite Integrals
1. ( ) 0aa f x dx =∫
2. ( ) ( )b aa bf x dx f x dx= −∫ ∫
3. ( ) ( ) ,b ba akf x dx k f x dx=∫ ∫ k a constant
4. [ ( ) ( )] ( ) ( )b b ba a af x g x dx f x dx g x dx± = ±∫ ∫ ∫
5. ( ) ( ) ( )b c ba a cf x dx f x dx f x dx= +∫ ∫ ∫
Calculus Chapter 6
6-24
In Problems 1 and 2, calculate the indicated Riemann sum nS for the function 2( ) 17 2 .f x x= −
1. Partition [–1, 9] into five subintervals of equal length, and for each subinterval
1[ , ],k kx x− let 1( ) / 2.k k kc x x−= +
9 ( 1) 10
25 5
x∆ − −= = =
11 1
02
c− += = 2
1 32
2c
+= = 33 5
42
c+= =
2(0) 17 2(0)
(0) 17
f
f
= −=
2(2) 17 2(2)
(2) 9
f
f
= −=
2(4) 17 2(4)
(4) 15
f
f
= −= −
45 7
62
c+= = 5
7 98
2c
+= =
2(6) 17 2(6)
(6) 55
f
f
= −= −
2(8) 17 2(8)
(8) 111
f
f
= −= −
5 1 2 3 4 5
5
5
5
( ) ( ) ( ) ( ) ( )
(17)(2) (9)(2) ( 15)(2) ( 55)(2) ( 111)(2)
34 18 30 110 222
310
S f c x f c x f c x f c x f c x
S
S
S
∆ ∆ ∆ ∆ ∆= ⋅ + ⋅ + ⋅ + ⋅ + ⋅= + + − + − + −= + − − −= −
2. Partition [–4, 8] into four subintervals of equal length, and for each subinterval
1[ , ],k kx x− let 1(2 ) / 3.k k kc x x−= +
8 ( 4) 12
34 4
x∆ − −= = =
12( 4) ( 1)
33
c− + −= = − 2
2( 1) 20
3c
− += =
2( 3) 17 2( 3)
( 3) 1
f
f
− = − −− = −
2(0) 17 2(0)
(0) 17
f
f
= −=
Calculus Chapter 6
6-25
32(2) 5
33
c+= = 4
2(5) 86
3c
+= =
2(3) 17 2(3)
(3) 1
f
f
= −= −
2(6) 17 2(6)
(6) 55
f
f
= −= −
4 1 2 3 4
4
4
4
( ) ( ) ( ) ( )
( 1)(3) (17)(3) ( 1)(3) ( 55)(3)
3 51 3 165
120
S f c x f c x f c x f c x
S
S
S
∆ ∆ ∆ ∆= ⋅ + ⋅ + ⋅ + ⋅= − + + − + −= − + − −= −
In Problems 3–7, calculate the definite integral, given that
50 12.5x dx =∫ 5 2
0125
3x dx =∫ 7 2
5218
3x dx =∫
3. 50 3x dx∫
5 50 0
50
3 3
3(12.5)
3 37.5
x dx x dx
x dx
=∫ ∫=
=∫
4. 5 20 (2 )x x dx+∫
5 5 52 20 0 0
5 20
(2 ) 2
1252(12.5)
3125
253
200(2 )
3
x x dx x dx x dx
x x dx
+ = +∫ ∫ ∫
= +
= +
+ =∫
Calculus Chapter 6
6-26
5. 7 20 2x dx∫
( )7 72 20 0
5 72 20 5
7 20
2 2
2
125 2182
3 3
3432
3
6862
3
x dx x dx
x dx x dx
x dx
=∫ ∫
= +∫ ∫
⎛ ⎞= +⎜ ⎟⎝ ⎠
⎛ ⎞= ⎜ ⎟⎝ ⎠
=∫
6. 05 7x dx∫
0 55 0
05
7 7
7(12.5)
7 87.5
x dx x dx
x dx
= −∫ ∫= −
= −∫
7. 7 27 ( 2 15)x x dx+ −∫
Since the upper and lower limits of integration are the same, the value of the integral
would be 7 27 ( 2 15) 0.x x dx+ − =∫
Calculus Chapter 6
6-27
Name ________________________________ Date ______________ Class ____________
Goal: To use the fundamental theorem of calculus to solve problems.
In Problems 1–7, evaluate the integrals.
1. 5 21 (6 2)x dx+∫
( )
53
5 21
15
3
1
3 3
5 21
6 2(6 2)
3 1
2 2
(2(5) 2(5)) (2(1) 2(1))
260 4
(6 2) 256
x xx dx
x x
x dx
⎛ ⎞+ = +∫ ⎜ ⎟
⎝ ⎠
= +
= + − += −
+ =∫
Section 6-5 The Fundamental Theorem of Calculus
Theorem 1: Fundamental Theorem of Calculus If f is a continuous function on [a, b], and F is any antiderivative of f, then
( ) ( ) ( )ba f x dx F b F a= −∫
Definition: Average Value of a Continuous Function f over [a, b]
1
( )ba f x dx
b a∫−
Calculus Chapter 6
6-28
2. 40 (7 )xe dx∫
44
0 0
4 0
4
40
(7 ) 7
7 7
7 7
(7 ) 375.19
x x
x
e dx e
e e
e
e dx
=∫
= −
= −
≈∫
3. 3 3 23 ( 2 8 7)x x x dx+ − +∫
Since the upper limit and lower limit of integration are the same value, the definite
integral value will be 3 3 23 ( 2 8 7) 0.x x x dx+ − + =∫
4. 2213
5
3dx
x∫
+
12
12
12
1 12 2
22 2213 13
22
12 13
22
13
2213
55( 3)
3
5( 3)
10( 3)
10(22 3) 10(13 3)
10(5) 10(4)
510
3
dx x dxx
x
x
dxx
−= +∫ ∫+
+=
= +
= + − += −
=∫+
Calculus Chapter 6
6-29
5. 1 2 3 60 2 (2 9)x x dx+∫
We will solve the problem using the substitution method of integration. Let
32 9,u x= + therefore 26 3 .du x dx du dx= ⇒ = Rewrite the original integral in terms of the variable u and solve.
1 12 3 6 60 0
17
0
13 7
0
3 7 3 7
1 2 3 60
12 (2 9)
3
21
(2 9)
21
(2(1) 9) (2(0) 9)
21 21
2 (2 9) 700,200.10
xx
x
x
x x dx u du
u
x
x x dx
==
=
=
+ =∫ ∫
=
+=
+ += −
+ ≈∫
6. 2
21 3 2
3 2 5
5 3
x xdx
x x x
− +∫
− + −
We will solve the problem using the substitution method of integration. Let
3 2 5 3,u x x x= − + − therefore 2(3 2 5) .du x x dx= − + Rewrite the original integral in terms of the variable u and solve.
22 21 13 2
2
1
23 2
1
3 2 3 2
221 3 2
3 2 5 1
5 3
ln
ln 5 3
ln (2) (2) 5(2) 3 ln (1) (1) 5(1) 3
ln 11 ln 2
3 2 51.70
5 3
xx
x
x
x xdx du
ux x x
u
x x x
x xdx
x x x
==
==
− + =∫ ∫− + −
=
= − + −
= − + − − − + −
= −
− + ≈∫− + −
Calculus Chapter 6
6-30
7. 2 21 3 6 1x x dx+∫
We will solve the problem using the substitution method of integration. Let
26 1,u x= − therefore 12 4(3 ) .du xdx du x dx= ⇒ = Rewrite the original integral in terms of the variable u and solve.
12
32
32
3 32 2
2 221 1
2
1
22
1
2 2
2 21
13 6 1
4
1 2
4 3
1(6 1)
6
1 1(6(2) 1) (6(1) 1)
6 6
125 343
6 6
3 6 1 17.75
xx
x
x
x x dx u du
u
x
x x dx
==
=
=
+ =∫ ∫
= ⋅
= +
= + − +
= −
+ ≈∫
In Problems 8 and 9, find the average value of the function over the given interval.
8. 3( ) 4 8 2;f x x x= − + [2,8]
( )( )
( )
8 3
2
84 2
2
4 2 4 2
1 1( ) (4 8 2)
8 21
4 261
((8) 4(8) 2(8)) ((2) 4(2) 2(2))61
3856 46642
b
af x dx x x dx
b a
x x x
= − +− −
= − +
= − + − − +
= −
=
∫ ∫
Calculus Chapter 6
6-31
9. 0.3( ) 2 xg x e−= [0,10]
We will solve the problem using the substitution method of integration. Let 0.3 ,u x= − therefore 0.3 .du dx= − Rewrite the original integral in terms of the variable
u and solve
( )( )
10 0.30
100
10
0
100.3
0
0.3(10) 0.3(0)
3
1 1( ) (2 )
10 02
( )(10)( 0.3)
2
3
2
3
2
32
13
0.63
b xa
x ux
xu
x
x
f x dx e dxb a
e du
e
e
e e
e
−
==
=
=
−
− −
−
=∫ ∫− −
∫−
⎛ ⎞= − ⎜ ⎟⎝ ⎠
⎛ ⎞− ⎜ ⎟⎝ ⎠
= − −
= − −
≈
10. The total cost (in dollars) of manufacturing x units of a product is ( ) 30,000 250 .C x x= +
a. Find the average cost per unit if 400 units are produced. [Hint: Recall that ( )C x is the average cost per unit.]
( )
( )
30,000 250( )
30,000( ) 250
C xC x
xx
C xx
C xx
=
+=
= +
30,000
(400) 250400
(400) 75 250
(400) 325
C
C
C
= +
= +
=
Therefore, the average cost for 400 units is $325 per unit.
Calculus Chapter 6
6-32
b. Find the average value of the cost over the interval [0, 400].
( )( )
82
4002
0
2 2
1 1( ) (30,000 250 )
400 0
130,000 125
400
1((30,000)(400) 125(400) ) ((30,000)(0) 125(0) )
4001
32,000,000 040080,000
ba f x dx x dx
b a
x x
= +∫ ∫− −⎛ ⎞= +⎜ ⎟⎝ ⎠
= + − +
= −
= Therefore, the average value of the cost will be $80,000.
11. A company manufactures a product and the research department produced the marginal cost function
'( ) 3004
xC x = − 0 800x≤ ≤
where '( )C x is in dollars and x is the number of units produced per month. Compute
the increase in cost going from a production level of 400 units per month to 800 units per month. Set up a definite integral and evaluate it.
800400
8002
400
2 2
'( ) (300 )4
3008
(800) (400)300(800) 300(400)
8 8
160,000 100,000
'( ) 60,000
ba
ba
xC x dx dx
xx
C x dx
= −∫ ∫
⎛ ⎞= −⎜ ⎟⎝ ⎠
⎛ ⎞ ⎛ ⎞= − − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= −
=∫
Therefore, the increase in cost going from 400 units per month to 800 units per month is $60,000.