Integration - Unit 5
Antiderivatives:
This
Chap 5: □ Sample Problem (by Marina Mendoza & Beni Atibalentja): (Medium) USING A “U” SUBSTITUTION:
e 2 1

0
x 3
dx
x6
SOLUTION:
e 2 1

0
x 3
dx
x6
With a “u” substitution:
u–6=x
u=x+6
du = dx
=

(u  6)  3
du
u
=

u 9
du
u

= (1 
=
9
)du
u
du
 1du - 9  u
= u – 9 ln u
e 2 1
e 2 1
= [(x + 6) ] 0
- 9 [ln |x + 6| ] 0
= (e2 + 1 + 6) – (0 + 6) – 9 (ln |e2 + 1 + 6|) + 9 (ln |0 + 6|)
= (e2 + 1) – 9 ln |
e2  7
|
6
Chap 5: □ Sample Problem (by Marina Mendoza & Beni Atibalentja): USING RIEMANN SUM:
PROBLEM: (Hard)
5
Find
 (4  4 x  x
2
SOLUTION:
2
) dx using Riemann Sum.
x 
52 3

n
n
ci  xi  a  ix  2 
n

3i
n
3i
3i 2  3
 n
 4  4(2  n )  (2  n )
lim
n 
i 1
3 n 
12i
12i 9i 2 
= lim
4
 2

4  8 
n n
n
n
n 
i 1 
3 n 
24i 9i 2 
16

 2

n i 1 
n
n 
= lim
n
3 n
24 n
9 n 2
16

i

 n
i
n  n 
n 2 i 1 
i 1
 i 1
3
24 n(n  1) 9 n(n  1)( 2n  1) 
= lim
16n 

 2


n  n
n
2
6
n

36(n  1) 9(n  1)( 2n  1) 

= lim 48 


n 
n
2n 2

= lim
= 48 + 36 + 9
= 93
Chap 5 : □ Sample problem (by Derek Caetano-Anolles & Ken Cebrian): Area of f(x) = sin x
Consider the graph of sinx from 0 to . Now find the area under the curve,
Area is found by regular integration.
A


0
sin x dx

 cos x 0  cos( )  cos(0)
 2 units2

Chap 5: □ Sample problem (by Emily Selen, Alex Chew, & Brian Atchley): This is a basic-level problem. It deals with
average value, and should be given to students when introducing this concept.
Find the average value of f ( x)  sin x 2 on the interval [1,3]
Solution: Average value is found using the formula
b
1
f ( x)dx
b  a a
Therefore the average value of f on [1,3] is:
3
1
sin x 2 dx
3  1 1
Use your calculator to solve and you get: f avg  .232
Chap 5: □ Sample problem (by Emily Selen, Alex Chew, & Brian Atchley): : This is an intermediate-level problem. In
order to complete it students must understand how integrals are used to find area, how to use a u-substitution, and the
power rule. This should be used to test understanding after a student learns these concepts.
Find the area of the region in the first quadrant bound by:
y  x 9  x 2 , the x-axis, and the line x=4.
Solution: To solve this problem we will set up an integral defining the area. Then, we will use a u-substitution and solve
from there using the power rule.
To find the area, we use the integral
4
2
x
9

x
dx

0
Now use a u-substitution u=9+x2 , then du=2xdx and xdx 
du
. For the limits, when x=0, u=9 and when x=4, u=25.
2
The new integral is
25

u
9
du
2
This can now be solved using the power rule.
1
2
25

9
25
 1 2 32 
u du    u 
2 3 9
1
3 
3
3



=  25 2  9 2 
=
98
3
Chap 5: □ Sample problem (by Emily Selen, Alex Chew, & Brian Atchley): This is an intermediate-level problem. It
involves relating functions, reading graphs, integration, area formulas, and understanding how to estimate an integral.
1
-2
-1
-1
1
2
3
4
-2
2
From the above graph, what is
 f (2 x)dx ?
1
Solution: To solve this problem we will first need to relate f(x) to f(2x). We can do this with the graph, using a usubstitution u=2x. Then, du = 2dx, and dx =
1
du . So, converting the limits we get:
2
2
4
1
f
(
2
x
)
dx

f (u )du

2 2
1
This new integral represents the area under the graph of f(x) from -2 to 4, which we can find with area formulas.
From -2 to 2, the graph forms a positive trapezoid. Its area A =

1
(base1  base 2)  height
2
1
(4  1) *1
2
5
=
2
The rest of a graph forms a triangle, with area A = ½*base*height
1
*2*2  2
2
4
1
1 5
1
So,  f (u )du  (  2) 
2 2
2 2
4
1
Answer:
4
=
1
□ Sample problem (by Fan Huang & Fernanda Mendez): Chap 5 Evaluate
2x 2
0 7 x  4 dx .
Check to see if function is continuous over the interval (0, 1).
Function is discontinuous when the denominator is equal to zero:
7x – 4 =0  x = 4/7. Integral is undefined - Fundamental Theorem of Calculus doesn’t apply because of
discontinuity at x = 4/7.
Chap 5: □ Sample problem (by Frances Ha, Lusiana Hadi, & Amity Xu): Temperature is given the as a function of time
in the following table.
Time (min)
0
1
2
3
4
5
6
Temp (C)
50
65
75
80
88
94
100
a. Use the trapezoidal approximation to find the average value of the temperature function from 0 to 6 min, and compare
this to the ordinary average of the temperature data.
Solution: Note that we can’t use the fundamental theorem of calculus since we don’t have a continuous function. To
find the average temperature using Trapezoidal Approximation with n  6 , use the formula
A
60
50  2(65)  2(75)  2(80)  2(88)  2(94)  100  954  6 = 79.5 C.
2
2(6)
find arith mean
b. Find the quadratic regression equation for the data.
Solution: Using your graphing calculator you should get y  .7024 x 2  12.1071x  51.6667 . Make sure you can do
this!
c. Find the average temperature using the equation found in question b.
(Using Trapezoidal Approximations, term-by-term integration and quadratic regression).


6
1 6
1 1
(-.7024x 2  12.1071x  51.6667)dx 
Solution: A 
 3 (.7024) x 3  12 (12.1071) x 2 51.6667 x

0
60
6
0
1

(477.3552  0) = 79.5592 C. Note that there may be other regressions that fit the data better than a quadratic
6
regression. The approximate average temperature we calculate depends upon the regression used, of course, since it
affects the integrand. One way to determine how well a regression fits your data is to look at the correlation coefficient
(the r value). Do in Excel.
Chap 5: □ Sample problem (by David Mesri & Jake Mathis): As with all subjects of math, there exists fundamental
theorems. Use the fundamental theorem of calculus and a simple u-substitution to solve this integral:
 /2
 cos4 x  sin 4 x  dx .
0
Solution: To begin with this problem, you must know the fundamental theorem of Calculus:
b
 f x  dx
 F b   F a  where F '  x   f  x 
a
 2
 cos4 x sin 4 x dx
0
u  sin 4 x 
*Make sure to use the chain rule
du  4 cos4 x 
1
4

 u du
* Must put a ¼ outside to compensate
for the 4 that will be placed inside the
integral to agree with our du
 u2 
 
 2

1
4

1
sin 4 x 0
8

1
sin 4 2  sin 0
8
2


1
0  0
8


 2


 cos4 x sin 4 x dx
 0
0
Chap 5: □ Sample problem (by Morgan Holbrook & Justin Parks): Integrate:
 ( x  4)( x  2)(3x  3)(2 x  5)dx
=  (3x  3x  18x  24)(2 x  5)dx
3
2
 (6 x
=
 21x 3  51x 2  42 x  120)dx
4
6 5 21 4
x 
x  17 x 3  21x 2  120 x  C
5
4
=

Chap 5: □ Sample problem (by Morgan Holbrook & Justin Parks): Integrate:

Solution:
( x  3)( x  2)
dx
1 3 1 2
x  x  6 x  517
3
2
=
1 3 1 2
x  x  6 x  517
2
u= 3

=

( x  3)( x  2)
dx
1 3 1 2
x  x  6 x  517
3
2
.
x2  x  6
dx
1 3 1 2
x  x  6 x  517
3
2
du = ( x2 - x - 6) dx
1
du
u
= 2u1/2 + C
2
=
1 3 1 2
x  x  6x  517  C
3
2
Chap 5: □ Sample problem (by Morgan Holbrook & Justin Parks): Find the area under f(x) = 4-x2 from x= -2 to
x= 2
Solution: Area = 
2
2
(4  x 2 )dx
2
1 3

 4 x  3 x 
2
=
8
8
16
16
(8  )  (  8  )
( )  ( )
3
3 = 3
3
=
32
= 3
2
Chap 5: □ Sample problem (by Morgan Holbrook & Justin Parks): a  t  2t  3 at start x = 0 and v= 0. Find
equation for position.
Solution:
x
v
1 3
t  t 2  3t
3
1 4 1 3 3 2
t  t  t
12
3
2
Chap 5: □ Sample problem (by Morgan Holbrook & Justin Parks):
A particle is moving with v = -5t2 + 20t –10 where
a. Find time when a = 0.
dv
  10t  20
Solution: dt
0 = -10t +20
t=2
b. Find displacement from time found from part a where at t = 0, x = 23
x   (  5t 2  20t  10)dx
Solution:
Should be dt
5
x   t 3  10t 2  10t  C
3
x 
C = 23
5
(8)  10(4)  10(2)  23
3
x
89
3
Chap 5: □ Sample problem (by Liz King & Katherine Wallig): This one is a bit like a treasure hunt, each time you take
the antiderivative, there is a clue waiting there for you so you can go on to the next antiderivative and eventually find the
treasure of what the original equation y is.
d3y
 24 x 4  3x  cos x
3
dx
y(0) = 4, y’(0) = -1, y’’(0) = 3
d3y
and plug in 0 for x, figure out what C is in order to get 3, then you
dx 3
SOLUTION: First take the antiderivative of
d2y
will have
and do the same thing there until you have y.
dx 2
 (24 x
4
3
2
3
 3x  cos x)  8 x  2 x  sin x  C
3
1
3
2
 (8x  2 x 2  sin x  3)  4 x  4 x 2  cos x  3x  C
 (4 x
2
 4x
1
2
y’’(0)=3 so C = 3
y’(0)= -1 so C = 0
 cos x  3x)  4 x 1  8 x  sin x  1.5 x 2  C y(0) = 4 so C = 4
So our final answer is y  4 x 1  8 x  sin x  1.5 x 2  4
Chap 5: □ Sample problem (by Liz King & Katherine Wallig): This is comprehensive problem for students who have
completed learning about end-point, trapezoid, and Simpson’s method approximations. The problem combines all of these
so that the student can compare their accuracy by using the second version of the fundamental theorem of Calculus and
percent error.
Given a curve with the equation f ( x)  3 sin
methods. Find the percent error for each.
x
, find the area under the curve on the interval [0,2 ] , using the following
2
a. the fundamental theorem of Calc (n=8)
Solution:

2
0
3 sin
x
x
2
0
dx  [3(2) cos ]02  6 cos
 cos  12 .
2
2
2
2
b. left-end point approximation (n=8)
5 3 7
, ,
,2 } area
4 2 4
4 2 4



3
5
3
7
[ f (0)  f ( )  f ( )  f ( )  f ( )  f ( )  f ( )  f ( )]

4
4
2
4
4
2
4
Solution: Partition: {0,
=
  3
,
,
, ,

12  11.84
 .01328  1.328 %.
(0 + 1.148 + 2.12 + 2.77 + 3 + 2.77 + 2.12 + 1.148) = 11.84. Percent Error:
12
4
c. trapezoid approximation (n=8)
{0,
  3
Solution: Partition:
,
,
4 2 4
, ,
5 3 7
, ,
,2 }
4 2 4
2  0


3
5
3
7
[ f (0)  2 f ( )  2 f ( )  2 f ( )  2 f ( )  2 f ( )  2 f ( )  2 f ( )  f (2 )]
4
2
4
4
2
4
area  2(8)

= 8 (0 + 2(1.148) + 2(2.12) + 2(2.77) + 2(3) + 2(2.77) + 2(2.12) + 2(1.148) + 0) = 11.84066. Percent Error
12  11.84
 .01328  1.328
12

%.
d. Simpson’s method (n=8)
5 3 7
, ,
,2 }
4 2 4
4 2 4


3
5
3
7
2  0
[ f (0)  4 f ( )  2 f ( )  4 f ( )  2 f ( )  4 f ( )  2 f ( )  4 f ( )  f (2 )]
area 
4
2
4
4
2
4
3(8)
Solution: Partition: {0,
=
  3
,
,
, ,

(0 + 4(1.148) + 2(2.12) + 4(2.77) + 2(3) + 4(2.77) + 2(2.12) + 4(1.148) + 0) = 11.99669. Percent Error =
12
12  11.99669
 2.754  10  4  .02754 %.
12
Chap 5: □ Sample problem (by Liz King & Katherine Wallig): u-sub This one looks like a monster, hard and
complicated, but if you look closely you should notice that what is in the second set of parenthesis is the derivative of
what is in the first parenthesis divided by 4, and luckily there is a 4 just waiting there for you to use! Just use a simple usubstitution to figure this one out.
 4( x
4
 2 x 2  4 x)( x 3  x  1)dx
Solution: In this one, let u equal the portion in the first parenthesis do du is equal to 4 times what is in the second
u  x 4  2x 2  4x
du  (4 x 3  4 x  4)dx
 udu 
u 2 ( x 4  2 x 2  4) 2

2
2
Chap 5: □ Sample problem (by Liz King & Katherine Wallig): This is a tricky u-substitution problem that utilizes all
the different techniques students have learned for solving integrals. Integrate:
2
x
x  12
dx
1
1
x
1
dx
du 
dx
2

2 x ,
x  6 . Then let u  x  6 . So x  (u  6) and
Solution: First, factor out 2 , so we get: 2
1 (u  6) 2 (2u  12)
du
2 xdu  dx . Therefore, 2(u + 6)du = dx. Now, let’s plug it in: 2 
u
. After factoring out a 2 we
(u  6) 3
u 3  18u 2  108u  18
du
du
 u

u
get:
. When expanded, it looks like this:
Next, split up the integral and
3
216
u
u 2 du   18udu   108du  
du
 9u 2  108u  216 ln | u | C

u
divide through by u:
Now integrate: 3
( x  6) 3
 9( x  6) 2  108( x  6)  216 ln( x  6)  C .
Almost done! All that’s left is to substitute x back in:
3
Chap 5: □ Sample problem (by Liz King & Katherine Wallig): Find the average value of f(x) = sin x over the interval
[0,2 ] .
2
1
1
1
sin xdx 
[ cos x]02 
( cos 2  cos 0)  0 .

2  0 0
2
2
Solution: f avg 
Chap 5: □ Sample problem (by Liz King & Katherine Wallig): This is a second-semester problem for students who
already have a strong grasp on the correspondence between area and integrals. It gives the student practice distinguishing
between total area and net area.
Find both the net area and the total area under the curve f ( x)  sin x over the interval [0,3 ] .
Solution: Total Area =


0
sin xdx  
2

3
sin xdx   sin xdx  [ cos x] 0  [ cos x] 2  [ cos x] 32  (1  1)  (1  1)  (1  1)
= 2 + 2 + 2 = 6. Net Area =
2

3
0
sin xdx  [ cos x]30  1  1  2 .
chap 5: □ Sample problem (by Merla Hübler & Lisa Portis): Intermediate fundamental theorem of calculus #2. If
x4
f ( x)   (t 3  t 2 ) dt what is f ’(x) ?
x
Solution:
This is basically a more complex application of the fundamental theorem, so it just has to be applied systematically.
Also, you have to split up the integration because of the two variable limits, a is a constant inserted for this purpose.
f ' x  
d x4 3 2
(t  t )dt
dx x
x4
d  a 3 2

3
2
f ' ( x) 
(
t

t
)
dt

a (t  t )dt 
dx  x
x4
d x
f ' ( x)     (t 3  t 2 )dt   (t 3  t 2 )dt 
a
dx  a

Remember that for the fundamental theorem, if the integration is f(x),
d [ f ( x )]
dx

d [ f ( x )]
dt
dt
 dx
, so you can substitute the
variable limit in for t, but you also have to multiply by the derivatives of the variable limits: x 0 and 4x 3 .
f ' ( x)  ( x 3  x 2 )( x 0 )  ( x12  x 8 )( 4 x 3 )
f ' ( x)   x 3  x 2  4 x15  4 x11
f ' ( x)  4 x15  4 x11  x 3  x 2
f ' ( x)  4x15+4x11-x3-x2
3
chap 5: □ Sample problem (by Merla Hübler & Lisa Portis): u substitutions Integrate:

x
2
cos x3dx .
0
[ u=x3 du=3x2dx ]
Solution:
3
1
3
3


 cos udu
0
x
2
 13 [sin x 3 ]0 
3
3
cos x dx =
0
 13 (sin   sin 0)
 13 (0)  0
chap 5: □ Sample problem (by Kevin Stanford & Mike Mitchner): antiderivative of the sine function as well as
solving definite integrals. basic Evaluate


2
0
sin xdx .
Solution:


2
0


2
0



0

2
0
 
sin xdx   cos    cos0
2
2
sin xdx  0   1
2
sin xdx  0  1  1
0

sin xdx   cos x
chap 5: Kevin Stanford & Mike Mitchner u-substitutions , derivative of the secant definite integrals. Intermediate
 /4
Evaluate

0
sec 2 x tan x .
Solution: In order to solve this problem it must first be noticed, that secxtanx is the derivative of secx. In this way, it is
seen that the integral is simple secx multiplied by its derivative. Once we notice this, we can proceed to make a usubstituion by making u = secx.
Let u = secx
And du = secxtanx
Therefore, u0  sec0  1
 
 
u   sec   2
4
And,  4 

1

1

1

1
1 2 2
u 1
2
2
2
1
1 2
udu 
2   1
2
2
2
1
1
udu  2   1
2
2
2
1 1
udu  1  
2 2
2
udu 
 
In finding this solution we changed the range of the integral because we used the range for u and not for x. We did so by
x

1
udu  u
2
4 . This problem can also be solved by first finding 
simply finding out what u would be at x = 0 and
and then plugging in secx for u. The solution will be the same no matter what method is used.
chap 5: Kevin Stanford & Mike Mitchner intermediate
average value Find the average value for this function.
Solution: To solve this problem we will use the formula favg
=
1 b
f  x dx . In our problem, a is represented by the
b  a a
start of the shaded regions of the functions at 0, and b is
represented by the end of the shaded region which is 2 .
After seeing this, the values of a, b, and f(x) can be entered
into the formula to find the solution.
2
1 b
1
1 2
1






2 cos x  x02
f
x
dx


2
sin
x

1
dx


2
sin
x

1
dx




a
0
0
ba
2  0
2
2
1
2 cos 2  2   2 cos0  0  1 21  2  2 1  1 2   1.

2
2
2
f avg 
chap 5: Kevin Stanford & Mike Mitchner
second fundamental theorem of Calculus
Evaluate:
d x
sin 3 5t  4 dt


3
dx
2
,
Solution: The fundamental theorem of Calculus states that
d x
f t dt  f x  . So therefore, in this example
dx a
d x
sin 3 5t  4dt  sin 3 5 x  4 . In this problem our value for a is -3, but this value could be any number and the


3
dx
final answer would still be the same.
chap 5: Kevin Stanford & Mike Mitchner trapezoid and Simpson methods Find


3
0
sin xdx in three different ways:
Solution Intro To solve this problem the formulas for the trapezoid method and Simpson’s method must be known. Also,
a partition must be created to be used in the formula. By partition we mean to break up the integral into the given number
of intervals, 6. In this case we have intervals of /18. Once we have the formulas and the partition, the first two parts of
the problem can be solved by following the formulas. The third part of the problem does not require a formula, but does
require the knowledge of the antiderivative of the sin function. If this is known the rest of the problem is a simple definite
integral.
a. Use the trapezoid method to approximate the integral with n = 6.
Solution: Trapezoid Approximation Formula:
{0,
Partition:

b
a
f x dx 
   2 5 
ba
 f a   2 f x1   ...  2 f xn1   f b 
2n
.
, , ,
,
, }
18 9 6 9 18 3 .
 0

3
 
 
 
 2 
 5 
  
f 0  2 f    2 f    2 f    2 f    2 f    f  

26 
 18 
9
6
 9 
 18 
 3 


5.716  0.499 .
 0  20.174  20.342  20.5  20.643  20.766  0.866 
36
36


3
0
sin xdx 
b. Use Simpson’s method to approximate the integral with n = 6.
Solution: Simpson’s approximation formula:
ba
 f x dx  3n  f a   4 f x   2 f x   4 f x   ...  2 f x   4 f x   f b
b
1
a
2
3
n2
n 1
Same Partition used as in part 1
 0

3
 
 
 
 2 
 5 
  
f 0  4 f    2 f    4 f    2 f    4 f    f  

0
3 6 
 18 
9
6
 9 
 18 
 3 


 0  40.174  20.342  40.5  20.643  40.766  0.866  8.596  0.500.
54
54


3
sin xdx 
c. Find the actual answer using the fundamental theorem.
Solution:

 /3
0
sin xdx   cos x
 /3
0
 
  cos    cos0 = -0.5 - (-1) = 0.5.
3
chap 5: Kevin Stanford & Mike Mitchner
Find the area under the curve on the closed interval [-1, 2] with the following function: f ( x)  2 x 2  x .
2
Solution:
 2x
2  x dx Let u  2  x , therefore x  2  u And
1
du  dx .   2(2  u )udu

  (4u  2u 2 )du .
0

Change limits into u limits:  (4u  2u 2 )du . Use the power rule to find
3
2 3 0
u ] 3 . Evaluate the integral
3
2
2
[2(0) 2  (0) 3 ]  [2(3) 2  (3) 3 ]  0.
3
3
the anti-derivative [2u 2 
chap 5: Kevin Stanford & Mike Mitchner Simpson’s method Using 6 equally spaced sub intervals approximate the area
under the curve from 0 to 6 of the following function using Simpson’s method for f ( x) 
1 3
x  ( x  2) 2 .
8
Solution: Simpson’s Method: b=6, a=0, n=6
ba
 f x dx  3n  f a   4 f x   2 f x   4 f x   ...  2 f x   4 f x   f b
b
1
a
2
n2
3
n 1
60
[ f (0)  4 f (1)  2 f (2)  4 f (3)  2 f (4)  4 f (5)  f (6)]
3(6)
1
1
 [4  4(1.125)  2(1)  4(4.375)  2(12)  4(24.625)  43]  [4  4.5  2  17.5  24  98.5  43]
3
3
1
 [193.5]  64.5 .
3

chap 5 Patrick McCall & Nathan Dornfeld: Fund. Thm #2 Find
d  e15 x cos 2 x


dx  15 csc 2 x
2x

0

f (t )dt  .

Solution: The derivative of a sum is the sum of the derivatives, so we need to find
d  e15 x cos 2 x  d


dx  15 csc 2 x  dx
which, by the fundamental theorem of calculus and the chain rule, is equivalent to
d  e cos 2 x 

  2 f ( x). The
dx  15 csc 2 x 
2x
 f (t )dt ,
0
15 x




1 d 15 x
1 d 15 x
e  2 sin 2 x cos 2 x . By the double angle
e cos 2 x sin 2 x =
30 dx
15 dx
1 d 15 x
formula for sine, this is equivalent to
e sin 4 x , and using the product and chain rules, this becomes
30 dx
1 15 x
2 15 x
1
e  4 cos 4 x  15e15 x  sin 4 x =
e cos 4 x  e15 x sin 4 x.
30
15
2
derivative can then be simplified to




chap 5: (by Raquel Roney, Nayeon Kang, & Ayush Dulguun): Fundamental Thm #1 Find c  0 such that
c
 (c  x)(c  x)dx  2
c
c
c
1
Solution: 2   (c  x)(c  x)dx =  (c 2  x 2 )dx = c 2 x  x 3
c
2
3
= c 3  (c 3 

3
c
c
1
3
1
3
= c 3  c 3  (c 2  c  (c)3
c
3
1 3
2
2
2
2
4
4
c ) = c 3  ( c 3 ) = c 3  c 3 = c 3 . Then, 2 = c 3  c  3 .
2
3
3
3
3
3
3
3


chap 5: (by Raquel Roney, Nayeon Kang, & Ayush Dulguun): proof by induction Prove that
= 1, 2, 3….





n
2
i1
 2 n 1 for n
i1
1
2
 211  2 0  1  2 1  21 1. Thus the formula works for the first natural

i1
number. To show that it holds for the rest of them, we must show that the formula working for an arbitrary
natural number, k, implies that the formula works for the next natural number, k + 1. That is, we assume the
Solution: Base case, n = 1:
formula holds for 
k:
k
2
i1
i1
 2 k 1. This is called the inductive hypothesis. When n = k + 1, the left side of
i1
k 1
k
i 1
i 1
i 1
i 1
 k 11
the formula can be written as  2   2  2
. Using our inductive hypothesis this is equivalent to
( k 1)1
(2  1)  2
k
 2 k  1  2 k  2  2 k  1  2 k 1  1. This last expression is exactly what the right side of
our formula gives when n = k + 1. Hence, by induction, the formula holds for all natural numbers.