1.
2.
3.
4.
5.
6.
7.
8.
R
R
R
R
R
R
R
R
Table of indefinite integrals
R
adx = ax + C, a is a constant, 9. tan xdx = − ln | cos x| + C,
R
2
10. cot xdx = ln | sin x| + C,
xdx = x2 + C,
R
n+1
xn dx = xn+1 + C, n 6= −1,
11. sec2 xdx = tan x + C,
R
1
dx = ln |x| + C,
12. csc2 xdx = − cot x + C,
x
R
ex dx = ex + C,
13. sec x tan xdx = sec x + C,
R
x
ax dx = lna a + C,
14. csc x cot x = − csc x + C,
R 1
sin xdx = − cos x + C,
15. √1−x
2 dx = arcsin x + C,
R 1
cos xdx = sin x + C,
16. 1+x2 dx = arctan x + C.
Definition of a definite integral
If f is a function defined on a closed interval [a, b], let P be a partition of [a, b] with partition
points x0 , x1 ,...,xn , where
a = x0 < x1 < x2 < ... < xn = b
Choose points x∗i ∈ [xi−1 , xi ] and let ∆xi = xi − xi−1 and kP k = max{∆xi }. Then the definite
integral of f from a to b is
Zb
f (x)dx = lim
kP k→0
a
n
X
f (x∗i )∆xi
i=1
if this limit exists. If the limit does exist, then f is called integrable on the interval [a, b].
In the notation
Rb
f (x)dx, f (x) is called the integrand and a and b are called the limits of
a
integration; a is the lower limit and b is the upper limit.
The procedure of calculating an integral is called integration.
Properties of the definite integral
Rb
1. cdx = c(b − a), where c is a constant.
a
2.
3.
4.
5.
Rb
a
Rb
a
Rb
a
Rb
a
cf (x)dx = c
Rb
f (x)dx, where c is a constant.
a
[f (x) ± g(x)]dx =
f (x)dx =
Rc
Rb
a
f (x)dx ±
f (x)dx +
a
Rb
Rb
g(x)dx.
a
f (x)dx, where a < c < b.
c
Ra
f (x)dx = − f (x)dx.
b
6. If f (x) ≥ 0 for a < x < b, then
Rb
a
f (x)dx ≥ 0.
7. If f (x) ≥ g(x) for a < x < b, then
Rb
a
f (x)dx ≥
1
Rb
a
g(x)dx.
8. If m ≤ f (x) ≤ M for a < x < b, then m(b − a) ≤
b
R
Rb
9. f (x)dx ≤ |f (x)|dx
a
Rb
a
f (x)dx ≤ M(b − a).
a
Section 6.4 The fundamental theorem of calculus.
Suppose f is continuous on [a, b].
1. If g(x) =
Rx
f (t)dt, then g ′ (x) = f (x).
a
2.
Rb
a
f (x)dx = F (b) − F (a) = F (x)|ba , where F is an antiderivative of f .
Example 1. Find the derivative of the function.
1. g(x) =
Rx
π
1
dt
1 + t4
SOLUTION. By the fundamental theorem of calculus, g ′(x) =
R4
√
t)8 dt
R4
√
Rx
√
SOLUTION. f (x) = x (2 + t)8 dt = − 4 (2 + t)8 dt.
2. f (x) =
x
(2 +
By the fundamental theorem of calculus, f ′ (x) = −(2 +
3. y(x) =
R 17
tan x
1
1 + x2
√
x)8
sin(t4 )dt
R tan x
sin(t4 )dt = − 17 sin(t4 )dt
Ru
Let u = tan x, then y(x) = − 17 sin(t4 )dt
SOLUTION. y(x) =
R 17
tan x
By the fundamental theorem of calculus, y ′ (x) = − sin(u4 )
du
= − sin(tan4 x) sec2 x
dx
Example 2. Evaluate the integral.
1.
Z6
√
1+ y
dy
y2
2
SOLUTION.
Z6
Z6
Z6
Z6
√
1+ y
√ −2
√
dy = (1 + y)y dy = (y −2 + y −2 y)dy = (y −2 + y −3/2 )dy
y2
2
2
=
2.
Z2
0
2
2
6 6
√
1
2 2
2
2
y −1 y −1/2 1
1
1
= − −√ =− −√ + +√ = −√ + 2
+
−1 −1/2 2
y
y 2
6
3
6 2
2
6
f (x)dx, where f (x) =
x4 0 ≤ x < 1
x5 1 ≤ x ≤ 2
2
SOLUTION.
R2
f (x)dx =
R1
f (x)dx +
0
0
1 26 1
1 64 1
107
+
− = +
− =
5
6
6
5
6
6
10
R2
f (x)dx =
R1
0
1
4
x dx +
R2
1
1
2
x6 x5 +
=
x dx =
5 0
6 1
5
Example 3. A particle moves along a line so that its velocity at time t is v(t) = t2 −2t−8.
1. Find the displacement of the particle during the time period 1 ≤ t ≤ 6.
6
6
3
6 2
R
R 2
t
2t
−
− 8t =
SOLUTION. Displacement= v(t)dt = (t − 2t − 8)dt = 3
2
1
1
1
3
6
− 36 − 48 − 1 + 1 + 8 = 10
3
3
3
2. Find the distance traveled during this time period.
R6
SOLUTION. Distance= |v(t)|dt
1
v(t) > 0 on (4, 6) and v(t) < 0 on 1, 4, therefore
R6
R
|v(t)|dt = −
6
3
2t2
t
−
− 8t 3
2
4
1
4
t3 2t2
− 2t − 8)dt = −
− 2t − 8)dt +
−
− 8t +
3
2
1
3
3
3
1
6
4
98
4
= − +42 +(4)(8)+ −1−8+ −62 −(6)(8)− +42 +(4)(8) =
3
3
3
3
3
lim41 (t2
R
lim64 (t2
3