Lesson 13
Solving Definite Integrals
How to find antiderivatives
We have three methods:
1. Basic formulas
2. Algebraic simplification
3. Substitution
Basic Formulas
If f(x) is…
x
n
k
cos(kx)
sin(kx)
ekx
1
x
ax
…then an antiderivative is…
x n +1 except if n = –1
kx (assuming the variable is x!)
1
n +1
sin(kx)/k
–cos(kx)/k
ekx /k
ln x
ax /ln(a)
Algebraic Simplification
2
1 3
(
x
!
1)(
x
+
1)
dx
=
x
!
1
dx
=
3 x ! x+C
"
"
! (3x + 1) dx =
2
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
!
!
x + x2
dx =
x
!
x (1 + x )
dx = ! 1 + x dx = x + 12 x 2 + C
x
Integrals by Substitution
Start with
Let u = g(x).
!
x sin(x 2 )dx
du
= g!(x) so du = g!(x) dx
dx
Let u =?
du
u = x 2 ! du = 2xdx
1
1
1
2
2
x
sin(
x
)
dx
=
sin(
x
)
2
x
dx
=
sin(
u
)
du
=
!
cos(
x
)+C
"
"
"
2
2
2
2
Antiderivative Practice
4t
t
2sin(3t)
!
e
+
4
dt Use basic formulas:
"
Problem 1
4t
t
2
1 4t
2
sin(3t)
!
e
+
4
dt
=
!
cos(3t)
!
+
3
4 e
"
1
ln(4 )
4t + C
z 2 + 2z
dz Simplify algebraically first, then integrate.
Problem 2 !
2
z
z2 + 2z
z2 2z
2
! z 2 dz = ! z 2 + z 2 dz = ! 1 + z dz = z + 2 ln z + C
Problem 3
!
6t
dt
2
4+t
Make a substitution: Let u=4+t2, so du=2tdt.
6t
2t
1
2
dt
=
3
dt
=
3
du
=
3ln
u
+
C
=
3ln
4
+
t
+C
! 4 + t2
! 4 + t2
!u
Antiderivative Practice
Problem 4
!
!
2
dy Make a substitution: u=ln(ky), so du=dy/y.
y ln(ky)
2
2 1
2
dy = !
dy = ! du = 2 ln u + C = 2 ln ln(ky ) + C
y ln(ky )
ln(ky ) y
u
Problem 5 Find the particular function F(x) such that F'(x) = x2 and
the graph of F(x) passes through (1, 2).
The general antiderivative is ! x 2 dx = 13 x 3 + C
3
Then to find C, we must have F(1) = 13 1 + C = 2
Thus, C = 5/3, and our function is F(x) = 13 x 3 + 53
Solving Definite Integrals
Theorem: (Fundamental Theorem I)
Or: If F is an antiderivative for f, then
Example
We determined using Simpson’s Rule :
Now use the fundamental theorem:
An antiderivative for f(x) = 3x + 5 is
So:
!
12
0
3x + 5 dx = F(12) - F(0) = 276 - 0 = 276
Example
!
3
2
x 3 dx
We have to
• find an antiderivative;
• evaluate at 3;
• evaluate at 2;
• subtract the results.
!2 x dx = x
3 3
1
4
4 3
2
= 14 3 " 14 2 =
4
This notation means:
evaluate the function at
3 and 2, and subtract the
results.
4
81
4
" 146 = 645 = 16.25
Don’t need to include “+ C” in our
antiderivative, because any
antiderivative will work.
Examples
"
!
2sin( x ) + 3x dx
0
)
!
0
3x 2
2sin( x ) + 3x dx = $2 cos x +
2
Alternate notation
= –1
!
0
=1
"
3! 2 #
= % $2 cos ! +
$ ($2 cos 0 + 0 )
&
2 (
'
"
3! 2 #
3! 2
= % $2( $1) +
$ ($2(1) + 0 ) = 4 +
&
2 (
2
'
1
"!2 s ds
!1
1
"!2 s ds = ln s
!1
!1
!2
= ln1 ! ln 2 =!! ln 2
Practice Examples
!
5
!
9
1
2
1
4
dx =
e
e
3 s ds =
!
9
2
3s ds = 2 s
1/ 2
3/ 2
9
2
()
=2 9
3/ 2
()
"2 2
3/ 2
= 54 " 4 2
ses + 1
"!2 s ds
!1
!1
1
s
s
= " e + ds = (e + ln s ) !2 = e!1 + ln1 ! (e!2 + ln 2)
!2
s
1 1
=!! ! 2 ! ln 2
e e
!1
Substitution in Definite Integrals
• We can use substitution in definite integrals.
• However, the limits are in terms of the original variable.
• We get two approaches:
– Solve an indefinite integral first
– Change the limits
Method I:
First solve an indefinite integral to find an antiderivative.
Then use that antiderivative to solve the definite integral.
Note: Do not say that a definite and an indefinite integral are equal
to each other! They can’t be.
Example
First: Solve an indefinite integral.
!
3t
3
dt =
2
t +4
2
!
!
2t
3
dt =
2
t +4
2
u = t2 + 4
3t
dt
2
t +4
!
du = 2t dt
1
du = 32 ln t 2 + 4 + C
u
Here’s an
antiderivative!
2t dt
becomes du
Pull out the 3,
and put in a 2.
Second: Use the antiderivative to solve the definite integral.
!
2
1
3t
dt =
2
t +4
(
3
2
ln t + 4
2
Here’s the antiderivative
we just found.
) (
2
1
=
3
2
) (
ln 8 "
3
2
)
ln 5 = 23 ln 85
Example
When discussing population growth, we worked backwards to find out
what we got from evaluating
P!(t )
Let’s find an antiderivative using substitution. "
dt
P (t )
"
P !(t)
dt =
P(t)
"
u = P(t)
du = P'(t) dt
1
du = ln u + C = ln P(t) + C
u
Of course, P(t) is always non-negative, so we don’t need absolute
values…
$ P(b) '
P !(t)
b
)
So we get: "a
dt = ln(P(t)) a = ln (P(b)) # ln (P(a)) = ln&
P(t)
% P(a) (
b
Method II: Convert the Limits
We start with x = a and x = b, and a substitution formula u = …
Just put a and b into the substitution formula and get new limits.
Note: You do not have to go back to x then!
2
3t
dt Start with the same substitution
Example !1 2
t +4
u = t2 + 4
du = 2t dt
8
3t
3 2 1
3 81
3
3
8
dt
=
2
t
dt
=
du
=
ln
u
=
ln
(
)
2
2
5
!1 t 2 + 4 2 !1 t 2 + 4
5
2 !5 u
When t = 1, u = 5.
becomes du
becomes u
When t = 2, u = 8.
What happens to t = 1?
And when t = 2,
2
u = t2 + 4 = 1^2 + 4 = 5. u = t + 4 = 2^2 + 4 = 8.
2
Example
!
8
1
)
8
1
y
y
( 5 y )dy
2
3
( 5 y )dy = )
3
2
8
1
u = 5y2
y = 1: u = 5
du = 10y dy
y = 8: u = 320
4
3
1
1
320
1 8
1
1 u
y (5 y ) dy = ) (5 y 2 )3 10 y dy = ) u 3 du =
10 1
10 5
10 4
3
Note that we can also do
this problem without u-sub
--try algebraic simplification
1
2 3
320
5
4
4
"
3 !
= % 320 3 # 5 3 & $ 163.5
40 '
(
!
8
1
y
(
3
5y
2
) dy = ! y (5y )
8
1
1
2 3
1
5
" 1 2%
8
dy = ! $ 53 y 3 ' y dy = 53 ! y 3 dy
1
1
#
&
8
1
3
5
+1
3
1
8
8
y
3
=5
= 53 (8 3 ( 13 ) ) 163.5
5
8
+1
3
1
8
!
1
e4 x
Practice Example
dx
1+ e
Method I: Firstly compute
0
4x
u = 1 + e4 x
du = 4e4 x dx
!
1
0
!
e4 x
1 + e4 x
e4 x
!
1+ e
4x
dx
1
"
2
1
" +1
2
1
4e4 x
1 1
1
1 u
dx = !
dx = !
du = ! u du =
+C
1
4 1 + e4 x
4
4
4
u
" +1
2
1 12
1 + e4 x
= u +C =
+C
2
2
e4 x
1 + e4 x
dx =
1
1
1
1
1
1 + e4 x |10 =
1 + e4 "
1 + e0 =
1 + e4 "
2
2
2
2
2
2
u = 1 + e4 x u(0) = 1 + e4*0 = 2,u(1) = 1 + e4*1 = 1 + e4
Method II:
!
1
0
e
4x
1 + e4 x
dx =
4x
1
"
2
1
" +1
2
1 1 4e
1 1+ e 1
1 1+ e
1 u
1+ e4
dx
=
du
=
u
du
=
|
2
4 !0 1 + e4 x
4 !2
4 !2
4 1
u
" +1
2
1 12 1+ e4
1 + e4
2
= u |2 =
"
2
2
2
4
4
Using Definite Integrals
We can now evaluate many of the integrals that we have been able
to set up.
Example
Find area between y = sin(x) and the
x–axis from x = 0 to x = π, and from
x = 0 to x = 2π.
The area from 0 to π is clearly:
#
!
0
!
sin( x ) dx = " cos( x ) = 1 + 1 = 2
0
The area from 0 to 2π is more complicated. We note that
But this is obviously not the area!
The area from 0 to 2π can be found by:
#
!
0
2!
(
!
)(
2!
sin( x ) dx " # sin( x ) dx = " cos( x ) 0 + cos( x ) !
!
)= 4
"
2!
0
sin( x ) dx =0
Summary
• Used the fundamental theorem to evaluate
definite integrals.
• Made substitutions in definite integrals
– By solving an indefinite integral first
– By changing the limits
• Used the fundamental theorem to evaluate
integrals which come from applications.