The FUNDAMENTAL Theorem of Calculus
(yay!)
Warm-up
Suppose a particle is traveling at velocity v (t) = t 2 from t = 1 to
t = 2. if the particle starts at y (0) = y0 ,
1. what is the function y (t) which gives the particles position as
a function of time (will have a y0 in it)?
2. how far does the particle travel from t = 1 to t = 2?
Compare your answer to the upper and lower estimates of the area
under the curve f (x) = x 2 from x = 1 to x = 2:
n ✓
X
i=1
Upper
◆ ✓ ◆
i 2
1
1+
⇤
n
n
n
10
100
1000
Upper
2.485
2.34835
2.33483
n 1✓
X
i=0
Lower
◆ ✓ ◆
i 2
1
1+
⇤
n
n
Lower
2.185
2.31835
2.33183
The Fundamental Theorem of Calculus
Theorem (the baby case)
If F (x) is any function satisfying
Z
Q. What is
A. F (x) =
So
Z 2
Z
x3
3
2
d
dx F (x)
= f (x), then
b
f (x)dx = F (b)
F (a)
a
x 2 dx?
1
+C
x 2 dx = F (2)
F (1) =
1
=
8
3
✓
23
+C
3
1
7
=
⇡ 2.333
3
3
◆
✓
13
+C
3
◆
The Fundamental Theorem of Calculus
Theorem (the baby case)
If F (x) is any function satisfying
Z
Q. What is
A. F (x) =
So
Z
2
1
Z
2
b
f (x)dx = F (x)
a
d
dx F (x)
b
x=a
= f (x), then
= F (b)
F (a)
x 2 dx?
1
x3
3
x3
x 2 dx =
3
=
8
3
2
=
x=1
✓
23
3
◆
✓
1
7
=
⇡ 2.333
3
3
13
3
◆
(same answer!)
Examples
Use the fundamental theorem of calculus,
Z b
f (x)dx = F (b) F (a)
a
to calculate
Z 3
1.
3x dx
2.
3.
4.
Z
Z
Z
2
1
x 3 dx
1
⇡
sin(x) dx
0
0
sin(x) dx
⇡
The Fundamental Theorem of Calculus
Theorem (the big case)
If F (x) is any function satisfying
Z
b(x)
f (t)dt = F (t)
a(x)
Q. What is
Z
ln(x)
d
dt F (t)
b(x)
t=a(x)
= f (t), then
= F (b(x))
F (a(x))
t 2 dt?
sin(x)
1
A. F (t) = t 3 .
3
So
Z
ln(x)
sin(x)
1
t 2 dt = t 3
3
ln(x)
=
t=sin(x)
✓
1
(ln(x))3
3
◆
✓
◆
1
(sin(x))3 .
3
Examples
Use the fundamental theorem of calculus,
Z b(x)
f (t)dt = F (b(x)) F (a(x))
a(x)
to calculate
Z cos(x)
1.
3t dt
sin(x)
2.
3.
Z
Z
5x 2 3
t 3 dt
x+1
0
sin(t) dt
arccos(x)
For reference, we calculated
f (t) = t 2
Notice:
✓
d 1
(ln(x))3
dx 3
Z
b(x)
f (t) dt where
a(x)
a(x) = sin(x)
1
(sin(x))3
3
◆
b(x) = ln(x).
1
(ln(x))2 cos(x)(sin(x))2
x
= b 0 (x)f (b(x)) a0 (x)f (a(x)).
=
In general:
Z
d
dx
b(x)
f (t) dt = b 0 (x)f (b(x))
a0 (x)f (a(x)).
a(x)
(Don’t even have to know F (t)!)
Why?
d
Calculate
dx
Z
sin(x)
2
e t dt.
tan(x)
Z
2
Answer: We can’t even calculate
e t dt!
Example:
2
(There is no elementary function F (t) which satisfies F 0 (t) = e t )
But we know
So
Z
sin(x)
Z
2
e t dt is a function. Call it F (t).
2
e t dt = F (sin(x))
F (tan(x)).
tan(x)
Therefore
d
dx
Z
sin(x)
tan(x)
2
e t dt =
d
(F (sin(x))
dx
= cos(x)F 0 (sin(x))
= cos(x)f (sin(x))
= cos(x)e (sin(x))
2
F (tan(x)))
sec2 (x)F 0 (tan(x))
sec2 (x)f (tan(x))
sec2 (x)e (tan(x))
2
Part 2: Integration by substitution
Warmup
Fill in the blank:
d
1. Since
cos(x 2 + 1) =
dx
Z
so
2. Since
,
dx = cos(x 2 + 1) + C .
d
ln | cos(x)| =
dx
,
so
(Example:
d 3
dx x dx
Z
= 3x 2 , so
dx = ln | cos(x)| + C .
R
3x 2 dx = x 3 + C .)
Undoing chain rule
In general:
d
f (g (x)) = f 0 (g (x)) ⇤ g 0 (x),
dx
Z
f 0 (g (x)) ⇤ g 0 (x)dx = f (g (x)) + C .
so
Example: Calculate the (extremely suggestively written) integral
Z
cos(x 3 + 5x 10) ⇤ (3x 2 + 5 ⇤ 1 + 0)dx = sin(x 3 + 5x 10) + C
Check:
d
dx
sin(x 3 + 5x
10) = cos(x 3 + 5x
10) ⇤ (3x 2 + 5)X
Less obvious chain rules.
Look for a buried function g (x) and it’s derivative g 0 (x) which can
be paired with dx:
Examples:
Z
cos(x)
dx =
sin(x) + 1
Z
Z p
p
1
2
x x + 1dx =
x 2 + 1 ⇤ 2x dx
2
Z
Z
1
⇤ cos(x) dx
sin(x) + 1
p
Z
p
cos( x)
1
p dx = cos( x) ⇤ p dx
2 x
2 x
Method of Substitution
Look for a buried function g (x) and it’s derivative g 0 (x) which can
be paired with dx.
Z
cos(x)
Example:
dx
sin(x) + 1
Let u = g (x).
Let u = sin(x) + 1
Calculate du.
du
dx
= cos(x) so du = cos(x)dx
Clear out all of the x’s,
replacing them with u’s.
Z
1
du
u
Calculate the new integral.
Substitute back into x’s.
Check
d
dx
R
1
u du
= ln |u| + C
ln |u| + C = ln | sin(x) + 1| + C
ln | sin(x) + 1| + C =
1
sin(x)+1
⇤ cos(x)X
Method of Substitution
Look for a buried function g (x) and it’s derivative g 0 (x) which can
be paired with dx.
Z p
Example:
x x 2 + 1 dx
Let u = x 2 + 1
Let u = g (x).
Calculate c ⇤ du.
du
dx
Clear out all of the x’s,
replacing them with u’s.
Z
Calculate the new integral.
1
2
Substitute back into x’s.
Check
d 1 2
dx 3 (x
= 2x so 12 du = x dx
p
R
u ⇤ 12 du
u 1/2 du =
1
2
2 3/2
3u
+C
= 13 (x 2 + 1)3/2 + C
+ 1)3/2 + C =
13 2
3 2 (x
+ 1)1/2 ⇤ 2xX
Give it a try:
Look for a buried function g (x) and it’s derivative g 0 (x) which can
be paired with dx.
Example:
Let u = g (x).
Calculate c ⇤ du.
Clear out all of the x’s,
replacing them with u’s.
Calculate the new integral.
Substitute back into x’s.
Z
p
e x
p dx
x
Extra practice: Integrals with substitution
(Check answers by taking a derivative and/or on Wolfram Alpha)
1.
2.
3.
Z
Z
Z
3x)4 dx
17.
Z
x
dx
1 + x4
5 dx
18.
Z
p
x5
dx
1 + x3
19.
Z
p
x
dx
1+x
20.
Z
p
1
x x4
1
21.
Z
p
x x
1 dx
(7
p
3x
Z
p
1
dx
4x + 3
5.
Z
p
3
6.
7.
Z
8.
Z
x3
dx
1 + x8
16.
4.
Z
Z
(2x + 9)5 dx
1
4x
dx
1
3)3/2
(2x
dx
4x
dx
+3
2x2
22.
x+1
dx
x2 + 2x 3
23.
24.
9.
Z
4x 5
dx
2x2 5x + 1
10.
Z
9x2 4x + 5
dx
3x3 2x2 + 5x + 1
11.
Z
p
x2
12.
Z
p
2x 1
dx
x2 x 1
13.
Z
p
dx
p
x+a+ x+b
14.
Z
p
15.
Z
x2
dx
1 + x6
2x + 3
+ 3x
dx
1
3x
2
p
5
dx
Z
Z
p
x) 1 + x dx
(1
p
x2
1 dx
p
x 3x
2 dx
x
25.
Z
(2x
26.
Z
dx
3 5x
27.
3x
Z
28.
29.
30.
Z
Z
Z
Z
dx
p
3)
p
x2
1 + x dx
sin 3x dx
cos(5 + 6x) dx
sin(5
3x) dx
3x + 5 dx
31.
32.
33.
Z
Z
Z
45.
Z
sin 2x
dx
a2 cos2 x + b2 sin2 x
sin x cos x dx
46.
Z
2 cos x 3 sin x
dx
3 cos x + 2 sin x
sin3 x cos x dx
47.
Z
1 + cos x
dx
(x + sin x)3
48.
Z
sin x
dx
(1 + cos x)2
49.
Z
x2 sin x3 dx
sin x
dx
sin x cos x
csc2 (2x + 5) dx
34.
Z
p
35.
Z
p
cos x
p
dx
x
36.
Z
sin(ax + b) cos(ax + b) dx
50.
Z
cos3 x dx
51.
Z
52.
Z
53.
Z
cos 2x
(sin x + cos x)2
54.
Z
cos x sin x
(1 + sin 2x)
sec2 x
dx
1 + tan x
55.
Z
x sin3 (x2 ) cos(x2 ) dx
sin x
dx
1 + cos x
56.
p
sin x
dx
2 + 3 cos x
Z
57.
2x sec3 (x2 + 3) tan(x2 + 3) dx
sin 2x
dx
a2 + b2 sin2 x
Z
58.
Z
sin 2x
dx
(a + b cos 2x)2
37.
38.
Z
Z
39.
Z
40.
Z
41.
Z
42.
43.
44.
Z
Z
Z
cos x sin x dx
(1/x2 ) cos(1/x) dx
2x sin(x2 + 1) dx
tan
p
x sec2
p
x
p
x
dx
1
dx
tan x
1
dx
cot x
sec2 x
1
tan2 x
dx