INTEGRALS
Essential Calculus, James STEWART

October 17, 2011

Essential Calculus, James STEWART ()

INTEGRALS

October 17, 2011

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Indefinite integrals

Recall: A function F is called an antiderivative of f on an interval I if
F (x) = f (x) for all x in I .

Essential Calculus, James STEWART ()

INTEGRALS

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Indefinite integrals

Recall: A function F is called an antiderivative of f on an interval I if
F (x) = f (x) for all x in I .

Definition
The family of all the antiderivative of f is called indefinite integral of f , denoted by f (x)dx, f (x)dx = F (x) mean F (x) = f (x)

Essential Calculus, James STEWART ()

INTEGRALS

October 17, 2011

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Indefinite integrals

Recall: A function F is called an antiderivative of f on an interval I if
F (x) = f (x) for all x in I .

Definition
The family of all the antiderivative of f is called indefinite integral of f , denoted by f (x)dx, f (x)dx = F (x) mean F (x) = f (x) b Remark: A definite integral a f (x)dx is a number whereas an indefinite integral f (x)dx is a function (or family of function).

Essential Calculus, James STEWART ()

INTEGRALS

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Indefinite integrals
Remark: The symbol is called an integral sign, f (x) is called the integrand and a and b are called the limits of integration; The dx simply indicates that the independent variable is x. The procedure of calculating an integral is called integration.

Essential Calculus, James STEWART ()

INTEGRALS

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Indefinite integrals
Remark: The symbol is called an integral sign, f (x) is called the integrand and a and b are called the limits of integration; The dx simply indicates that the independent variable is x. The procedure of calculating an integral is called integration.

Theorem
If F (x) is an antiderivative of f then arbitrary constant.

Essential Calculus, James STEWART ()

f (x)dx = F (x) + C , where C is an

INTEGRALS

October 17, 2011

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Indefinite integrals
Remark: The symbol is called an integral sign, f (x) is called the integrand and a and b are called the limits of integration; The dx simply indicates that the independent variable is x. The procedure of calculating an integral is called integration.

Theorem
If F (x) is an antiderivative of f then arbitrary constant.

f (x)dx = F (x) + C , where C is an

Property
( f (x)dx) = f (x).
F (x)dx =

dF = F (x) + C .

(αf (x) + βg (x))dx = α

Essential Calculus, James STEWART ()

f (x)dx + β

INTEGRALS

g (x)dx, α, β are constant.

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Indefinite integrals
0dx = C , dx = x + C . dx = ln |x| + C x x α+1 x α dx =
+ C , α ∈ {−1, 0}. α+1 ax ax dx =
+ C. ln a sin xdx = − cos x + C , cos xdx = sinx + C tan xdx = − ln | cos x| + C , cot xdx = − ln | sin x| + C dx π x dx x = ln tan
= ln tan
+
+ C,
+C
sin x
2
cos x
2
4 dx dx
= tan x + C ,
= − cot x + C
2x
cos sin2 x

Essential Calculus, James STEWART ()

INTEGRALS

October 17, 2011

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Indefinite integrals
Expansion method: n n

αi fi (x)dx = i=1 αi

fi (x)dx,

i=1

αi are constants.
Change variable method: f (x)dx = F (x) + C ⇐⇒

Essential Calculus, James STEWART ()

f (x(t))x (t)dt = F (x(t) + C

INTEGRALS

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Indefinite integrals
Expansion method: n n

αi fi (x)dx = i=1 αi

fi (x)dx,

i=1

αi are constants.
Change variable method: f (x)dx = F (x) + C ⇐⇒

f (x(t))x (t)dt = F (x(t) + C

Theorem (The substitution rule)
If u = g (x) is a differentiable function whose range is an interval I and f is continuous on I , then f (g (x))g (x)dx =
Essential Calculus, James STEWART ()

INTEGRALS

f (u)du
October 17, 2011

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The substitution rule
Example

Example
Find I =

1

dx
,
(3 − 5x)2

dx
,
sin x

√
1+ x
√
1− x

Let u = 3 − 5x

Essential Calculus, James STEWART ()

INTEGRALS

October 17, 2011

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The substitution rule
Example

Example
Find I =

dx
,
(3 − 5x)2

dx
,
sin x

√
1+ x
√
1− x

1

Let u = 3 − 5x

2

1
Compute dx by using du = u (x)dx, dx = − 5 du

Essential Calculus, James STEWART ()

INTEGRALS

October 17, 2011

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The substitution rule
Example

Example
Find I =

1
2

3

dx
,
(3 − 5x)2

dx
,
sin x

√
1+ x
√
1− x

Let u = 3 − 5x
1
Compute dx by using du = u (x)dx, dx = − 5 du
1 du
1
1
Find the new integral: I = −
=
=
2
5 u
5u
5(3 − 5x)

Essential Calculus, James STEWART ()

INTEGRALS

October 17, 2011

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Indefinite integrals

1

Antiderivative of rational functions f (x) =

Essential Calculus, James STEWART ()

INTEGRALS

P(x)
:
Q(x)

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Indefinite integrals

1

Antiderivative of rational functions f (x) =
Write f (x) = T (x) +

P(x)
:
Q(x)

R(x)
, deg R(x) < deg Q(x)
Q(x)

Example
Calculate

x 4 − 2x 2 + 4x + 1 dx and x3 − x2 − x + 1

Essential Calculus, James STEWART ()

x 5 + 3x 3 + x 2 + 2x + 2 dx (x − 1)2 (x 2 + x + 1)2

INTEGRALS

October 17, 2011

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Indefinite integrals

1

Antiderivative of rational functions f (x) =

P(x)
:
Q(x)

R(x)
, deg R(x) < deg Q(x)
Q(x)
k αi 2 βi i=1h (x + pi x + qi ) i=1 (x − ci )

Write f (x) = T (x) +
Write Q(x) = an

Example
Calculate

x 4 − 2x 2 + 4x + 1 dx and x3 − x2 − x + 1

Essential Calculus, James STEWART ()

x 5 + 3x 3 + x 2 + 2x + 2 dx (x − 1)2 (x 2 + x + 1)2

INTEGRALS

October 17, 2011

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Indefinite integrals

1

Antiderivative of rational functions f (x) =

P(x)
:
Q(x)

R(x)
, deg R(x) < deg Q(x)
Q(x)
Write Q(x) = an k (x − ci )αi i=1h (x 2 + pi x + qi )βi i=1 R(x)
A1
Aαi −1
Aαi
Write
=
+ ··· +
+
+ αi −1
Q(x)
x − ci
(x − ci )
(x − ci )αi
Bβ −1 x + Cβi −1
Bβ x + Cβi
B1 x + C1
+ ··· + 2 i
+ 2 i
2+p x +q βi −1 x (x + pi x + qi )
(x + pi x + qi )βi i i

Write f (x) = T (x) +

Example
Calculate

x 4 − 2x 2 + 4x + 1 dx and x3 − x2 − x + 1

Essential Calculus, James STEWART ()

x 5 + 3x 3 + x 2 + 2x + 2 dx (x − 1)2 (x 2 + x + 1)2

INTEGRALS

October 17, 2011

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Indefinite integrals

Example sin x + sin3 x
1 + cos x dx, 2. dx, 3. sin x − 1
2 cos2 x − 1 cos3 x + cos5 x dx , 4. and 5. sinm x cosn xdx, m, n ∈ N∗
2
4 sin x cos x sin x + sin x

Calculate 1.

Essential Calculus, James STEWART ()

INTEGRALS

October 17, 2011

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Indefinite integrals

Example sin x + sin3 x
1 + cos x dx, 2. dx, 3. sin x − 1
2 cos2 x − 1 cos3 x + cos5 x dx , 4. and 5. sinm x cosn xdx, m, n ∈ N∗
2
4 sin x cos x sin x + sin x

Calculate 1.

Hint: 5. If m odd, we let t = cos x.
If n odd, we let t = sin x.
If m, n are both even then we use the trigonometric formulas

Essential Calculus, James STEWART ()

INTEGRALS

October 17, 2011

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Indefinite integrals

Antiderivative of some trigonometric and irrational functions Example: x 1 f (x) = R(sin x, cos x): Change variable t = tan and use
2
2t
1 − t2 sin x =
, cos x =
2
1+t
1 + t2 ax + b n ax + b
2 f (x) = R(x,
) then we let t = n cx + d cx + d

Essential Calculus, James STEWART ()

INTEGRALS

October 17, 2011

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Areas and distances
The area problem

Essential Calculus, James STEWART ()

INTEGRALS

October 17, 2011

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Areas and distances
The area problem

Essential Calculus, James STEWART ()

INTEGRALS

October 17, 2011

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Areas and distances
The area problem

Example

Ln ≤ A ≤ Rn

Essential Calculus, James STEWART ()

INTEGRALS

October 17, 2011

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Areas and distances
The area problem

Example

Ln ≤ A ≤ Rn
A = lim Ln = lim Rn n→∞ n→∞

where Ln and Rn is the sum of the areas of the n rectangles by using left and right enpoints.

Essential Calculus, James STEWART ()

INTEGRALS

October 17, 2011

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Areas and distances
The area problem

Question: How to compute the area in the more general case?
1

Subdivide S into n strips S1 , S2, · · · , Sn of equal width.

Essential Calculus, James STEWART ()

INTEGRALS

October 17, 2011

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Areas and distances
The area problem

Question: How to compute the area in the more general case?
1

Subdivide S into n strips S1 , S2, · · · , Sn of equal width.

2

The width of each of the n strips is ∆x =

Essential Calculus, James STEWART ()

INTEGRALS

b−a n October 17, 2011

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Areas and distances
The area problem

Question: How to compute the area in the more general case?
1

2
3

Subdivide S into n strips S1 , S2, · · · , Sn of equal width.

b−a n Compute Rn = f (x1 )∆x + f (x1 )∆x + · · · + f (xn )∆x, where the right endpoints xi = a + i∆x.
The width of each of the n strips is ∆x =

Essential Calculus, James STEWART ()

INTEGRALS

October 17, 2011

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Areas and distances
The area problem

Definition
The area A of the region S that lies under the graph of the continuous function f is the limit of the sum of the areas of approximating rectangles:
A = lim Rn = lim f (x1 )∆x + f (x2 )∆x + · · · + f (xn )∆x n→∞ Essential Calculus, James STEWART ()

n→∞

INTEGRALS

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Areas and distances
The area problem

Definition
The area A of the region S that lies under the graph of the continuous function f is the limit of the sum of the areas of approximating rectangles:
A = lim Rn = lim f (x1 )∆x + f (x2 )∆x + · · · + f (xn )∆x n→∞ n→∞

Similarly, we have:
A = lim Ln = lim f (x1 )∆x + f (x2 )∆x + · · · + f (xn )∆x n→∞ Essential Calculus, James STEWART ()

n→∞

INTEGRALS

October 17, 2011

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Areas and distances
The area problem

∗ ∗
∗
Let x1 , x2 , ..., xn are the sample points

∗
∗
∗
A = lim f (x1 )∆x + f (x2 )∆x + · · · + f (xn )∆x n→∞ Essential Calculus, James STEWART ()

INTEGRALS

October 17, 2011

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Areas and distances
The area problem

Notation: We use sigma notation to write sums with many terms more compactly: n

f (xi )∆x = f (x1 )∆x + f (x2 )∆x + · · · + f (xn )∆x i=1 Essential Calculus, James STEWART ()

INTEGRALS

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Areas and distances
The area problem

Notation: We use sigma notation to write sums with many terms more compactly: n

f (xi )∆x = f (x1 )∆x + f (x2 )∆x + · · · + f (xn )∆x i=1 So

n

A = lim

n→∞

f (xi )∆x i=1 n

= lim

n→∞

f (xi−1 )∆x i=1 n

f (xi∗ )∆x

= lim

n→∞

Essential Calculus, James STEWART ()

i=1

INTEGRALS

October 17, 2011

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Areas and distances
The area problem

Notation: We use sigma notation to write sums with many terms more compactly: n

f (xi )∆x = f (x1 )∆x + f (x2 )∆x + · · · + f (xn )∆x i=1 So

n

A = lim

n→∞

f (xi )∆x i=1 n

= lim

n→∞

f (xi−1 )∆x i=1 n

f (xi∗ )∆x

= lim

n→∞

i=1

In some cases, it is difficult to evaluate the limit by using right and left endpoints. Essential Calculus, James STEWART ()

INTEGRALS

October 17, 2011

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Areas and distances
The area problem

Notation: If we divide [a, b] into n smaller subintervals (not be possible the same width) (partition P), n ∗
∗
∗ f (xi∗ )∆xi = f (x1 )∆x1 + f (x2 )∆x2 + · · · + f (xn )∆xn i=1 is called a Riemann sum associated with a partition P.

Essential Calculus, James STEWART ()

INTEGRALS

October 17, 2011

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Areas and distances
The distance problem

Problem: Find the distance traveled by an object during a certain time period if the velocity of the object is known at all times.

Essential Calculus, James STEWART ()

INTEGRALS

October 17, 2011

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Areas and distances
The distance problem

Problem: Find the distance traveled by an object during a certain time period if the velocity of the object is known at all times.
→ distance = velocity × time

Essential Calculus, James STEWART ()

INTEGRALS

October 17, 2011

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Areas and distances
The distance problem

Problem: Find the distance traveled by an object during a certain time period if the velocity of the object is known at all times.
→ distance = velocity × time
Question: How to find the distance if the velocity varies?

Essential Calculus, James STEWART ()

INTEGRALS

October 17, 2011

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Areas and distances
The distance problem

Problem: Find the distance traveled by an object during a certain time period if the velocity of the object is known at all times.
→ distance = velocity × time
Question: How to find the distance if the velocity varies?
1

Suppose that the velocity v = f (t), a ≤ t ≤ b and f (t) ≥ 0

2

Take velocity reading at time t0 = a, t1 , ..., tn = b b−a The time between consecutive reading ∆t = n The distance

3
4

n

d = lim

n→∞

Essential Calculus, James STEWART ()

f (ti )∆t i=1 INTEGRALS

October 17, 2011

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The definite integral
Definition (Definite Integral)
If f is a function defined for a ≤ x ≤ b, we divide the interval [a, b] into n b−a . We let x0 = a, x1 , x2 , ..., xn = b subintervals of equal width ∆x = n ∗ ∗
∗
be the endpoints of these subintervals and we let x1 , x2 , ...xn be any
∗ lies in the ith subinterval sample points in these subintervals, so xi
[xi−1 , xi ]. Then the definite integral of f from a to b is n b

f (x)dx = a f (xi∗ )∆xi

lim

max ∆xi →0

i=1

provided that this limit exists and gives the same value for all possible choices of sample points. If it does exist, we say that f is integrable on
[a, b].

Essential Calculus, James STEWART ()

INTEGRALS

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The definite integral

1

The definite integral

Essential Calculus, James STEWART ()

b a f (x)dx is a number; it does not depend on x.

INTEGRALS

October 17, 2011

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The definite integral

1
2

The definite integral
The sum

b a n
∗
i=1 f (xi )∆x

Essential Calculus, James STEWART ()

f (x)dx is a number; it does not depend on x. is called a a Riemann sum.

INTEGRALS

October 17, 2011

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The definite integral

Essential Calculus, James STEWART ()

INTEGRALS

October 17, 2011

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The definite integral

Essential Calculus, James STEWART ()

INTEGRALS

October 17, 2011

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The definite integral b f (x)dx = A1 − A2 a where A1 and A2 is the area of the region above and below the x-axis and below the graph of f .

Essential Calculus, James STEWART ()

INTEGRALS

October 17, 2011

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The definite integral b f (x)dx = A1 − A2 a where A1 and A2 is the area of the region above and below the x-axis and below the graph of f .

Theorem
If f is continuous on [a, b], or if f has only a finite number of jump discontinuities, then f is integrable on [a, b]; that is, the definite integral b a f (x)dx exists

Essential Calculus, James STEWART ()

INTEGRALS

October 17, 2011

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The definite integral b f (x)dx = A1 − A2 a where A1 and A2 is the area of the region above and below the x-axis and below the graph of f .

Theorem
If f is continuous on [a, b], or if f has only a finite number of jump discontinuities, then f is integrable on [a, b]; that is, the definite integral b a f (x)dx exists

Theorem
If f is integrable on [a, b], then n b

f (x)dx = lim a Essential Calculus, James STEWART ()

n→∞

INTEGRALS

f (xi )∆x i=1 October 17, 2011

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The definite integral
Evaluating integrals

→ By taking the sample endpoint and remark that n n(n + 1)
1
i=
2
i=1

Essential Calculus, James STEWART ()

INTEGRALS

October 17, 2011

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The definite integral
Evaluating integrals

→ By taking the sample endpoint and remark that n n(n + 1)
1
i=
2
i=1 n n(n + 1)(2n + 1)
2
i2 =
6
i=1

→ By interpreting the integral in terms of areas.
Essential Calculus, James STEWART ()

INTEGRALS

October 17, 2011

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The definite integral
Evaluating integrals

→ By taking the sample endpoint and remark that n n(n + 1)
1
i=
2
i=1 n n(n + 1)(2n + 1)
2
i2 =
6
i=1 n 3

i=1

i3 =

n(n + 1)
2

2

→ By interpreting the integral in terms of areas.
Essential Calculus, James STEWART ()

INTEGRALS

October 17, 2011

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The definite integral
Evaluating integrals

→ By taking the sample endpoint and remark that n n(n + 1)
1
i=
2
i=1 n n(n + 1)(2n + 1)
2
i2 =
6
i=1 n 3

i=1 n i3 =

n(n + 1)
2
n

c = nc,

4

i=1

2

n

cai = c i=1 ai i=1 → By interpreting the integral in terms of areas.
Essential Calculus, James STEWART ()

INTEGRALS

October 17, 2011

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The definite integral
Evaluating integrals

→ By taking the sample endpoint and remark that n n(n + 1)
1
i=
2
i=1 n n(n + 1)(2n + 1)
2
i2 =
6
i=1 n i3 =

3

i=1 n n(n + 1)
2
n

c = nc,

4

i=1 n 5

n

cai = c i=1 n

(ai ± bi ) = i=1 2

ai i=1 n

ai ± i=1 bi i=1 → By interpreting the integral in terms of areas.
Essential Calculus, James STEWART ()

INTEGRALS

October 17, 2011

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The definite integral
Example
3

1. Evaluate 0 (x 3 − 6x)dx by evaluting the Riemann sum for f (x) = x 3 − 6x with n = 6.
1
2. Evaluate 0 (x − 1)dx by interpreting it in terms of areas.

Essential Calculus, James STEWART ()

INTEGRALS

October 17, 2011

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The definite integral
Example
3

1. Evaluate 0 (x 3 − 6x)dx by evaluting the Riemann sum for f (x) = x 3 − 6x with n = 6.
1
2. Evaluate 0 (x − 1)dx by interpreting it in terms of areas.

Essential Calculus, James STEWART ()

INTEGRALS

October 17, 2011

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The definite integral
Any Riemann sum is an approximation to an integral, if we use midpoints then Definition (The midpoint rule) n b

f (x)dx ≈ a where ∆x =

f (¯i )∆x, x i=1

b−a
1
and xi = (xi−1 + xi ) = midpoint of [xi−1 , xi ].
¯
n
2

Essential Calculus, James STEWART ()

INTEGRALS

October 17, 2011

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The definite integral
Any Riemann sum is an approximation to an integral, if we use midpoints then Definition (The midpoint rule) n b

f (x)dx ≈ a where ∆x =

f (¯i )∆x, x i=1

b−a
1
and xi = (xi−1 + xi ) = midpoint of [xi−1 , xi ].
¯
n
2

Example
With n = 5, we have
2
1

1 dx ≈ ∆x[f (1.1) + f (1.3) + f (1.5) + f (1.7) + f (1.9)] ≈ 0.691908. x Essential Calculus, James STEWART ()

INTEGRALS

October 17, 2011

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The definite integral
Properties of the definite integral

1

a b f (x)dx

=−

b a Essential Calculus, James STEWART ()

f (x)dx;

a a f (x)dx

INTEGRALS

= 0.

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The definite integral
Properties of the definite integral

1
2

a b a b f (x)dx = − a f (x)dx; a f (x)dx = 0. b b b a [f (x) ± g (x)]dx = a f (x)dx ± a g (x)dx.

Essential Calculus, James STEWART ()

INTEGRALS

October 17, 2011

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The definite integral
Properties of the definite integral

1
2
3

a b a b f (x)dx = − a f (x)dx; a f (x)dx = 0. b b b a [f (x) ± g (x)]dx = a f (x)dx ± a g (x)dx. b b b a cf (x)dx = c a f (x)dx, a cdx = c(b − a),

where c is a constant.

Essential Calculus, James STEWART ()

INTEGRALS

October 17, 2011

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The definite integral
Properties of the definite integral

1
2
3

a b a b f (x)dx = − a f (x)dx; a f (x)dx = 0. b b b a [f (x) ± g (x)]dx = a f (x)dx ± a g (x)dx. b b b a cf (x)dx = c a f (x)dx, a cdx = c(b − a),

where c is a constant.
4

c a f (x)dx +

b c f (x)dx =

Essential Calculus, James STEWART ()

b a f (x)dx

INTEGRALS

October 17, 2011

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The definite integral
Properties of the definite integral

1
2
3

a b a b f (x)dx = − a f (x)dx; a f (x)dx = 0. b b b a [f (x) ± g (x)]dx = a f (x)dx ± a g (x)dx. b b b a cf (x)dx = c a f (x)dx, a cdx = c(b − a),

where c is a constant.
4
5

c a f (x)dx +

b c f (x)dx =

b a f (x)dx

If f (x) ≥ 0 for a ≤ x ≤ b then

Essential Calculus, James STEWART ()

b a f (x)dx ≥ 0.

INTEGRALS

October 17, 2011

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The definite integral
Properties of the definite integral

1
2
3

a b a b f (x)dx = − a f (x)dx; a f (x)dx = 0. b b b a [f (x) ± g (x)]dx = a f (x)dx ± a g (x)dx. b b b a cf (x)dx = c a f (x)dx, a cdx = c(b − a),

where c is a constant.
4

c a f (x)dx +

b c f (x)dx =

b a f (x)dx b a

5

If f (x) ≥ 0 for a ≤ x ≤ b then

6

If f (x) ≥ g (x) for a ≤ x ≤ b then

Essential Calculus, James STEWART ()

f (x)dx ≥ 0. b a

INTEGRALS

f (x)dx ≥

b a g (x)dx.

October 17, 2011

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The definite integral
Properties of the definite integral

1
2
3

a b a b f (x)dx = − a f (x)dx; a f (x)dx = 0. b b b a [f (x) ± g (x)]dx = a f (x)dx ± a g (x)dx. b b b a cf (x)dx = c a f (x)dx, a cdx = c(b − a),

where c is a constant.
4

c a f (x)dx +

b c f (x)dx =

b a f (x)dx b a

5

If f (x) ≥ 0 for a ≤ x ≤ b then

f (x)dx ≥ 0.

6

If f (x) ≥ g (x) for a ≤ x ≤ b then

7

If m ≤ f (x) ≤ M for a ≤ x ≤ b then

b a f (x)dx ≥

b a g (x)dx.

b

m(b − a) ≤

f (x)dx ≤ M(b − a). a Essential Calculus, James STEWART ()

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Evaluating definite integrals
Theorem (Evaluation theorem)
If f is continuous on [a, b] then b f (x)dx = F (b) − F (a) a where F is any antiderivative of f , that is, F = f .

Example
Evaluate

4
1

1 dx and x2 Essential Calculus, James STEWART ()

b
0

sin xdx, where 0 ≤ b ≤ π/2.

INTEGRALS

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Evaluating definite integrals
Theorem (Evaluation theorem)
If f is continuous on [a, b] then b f (x)dx = F (b) − F (a) a where F is any antiderivative of f , that is, F = f .

Example
Evaluate

4
1

1 dx and x2 b
0

4
1

Essential Calculus, James STEWART ()

sin xdx, where 0 ≤ b ≤ π/2.

1
3
dx = F (4) − F (1) =
2
x
4

INTEGRALS

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Evaluating definite integrals
Theorem (Evaluation theorem)
If f is continuous on [a, b] then b f (x)dx = F (b) − F (a) a where F is any antiderivative of f , that is, F = f .

Example
Evaluate

4
1

1 dx and x2 b
0

4
1

sin xdx, where 0 ≤ b ≤ π/2.

1
3
dx = F (4) − F (1) =
2
x
4

b

b

sin xdx = (− cos x)
0
Essential Calculus, James STEWART ()

INTEGRALS

0

= 1 − cos b
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Fundamental theorem of calculus
Theorem (Fundamental theorem)
If f is continuous on [a, b] then the function g defined by x g (x) =

f (t)dt,

a≤x ≤b

a

is an antiderivative of f , that is, g (x) = f (x) for a < x < b.

Essential Calculus, James STEWART ()

INTEGRALS

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Fundamental theorem of calculus
Theorem (Fundamental theorem)
If f is continuous on [a, b] then the function g defined by x g (x) =

f (t)dt,

a≤x ≤b

a

is an antiderivative of f , that is, g (x) = f (x) for a < x < b.

Theorem (Fundamental theorem of calculus)
Suppose f is continuous function on [a, b].
If g (x) = b a

x a f (t)dt then g (x) = f (x).

f (x)dx = F (b) − F (a), where F = f .

Essential Calculus, James STEWART ()

INTEGRALS

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Fundamental theorem of calculus
Example

x√
Find the derivative of g (x) = 0 3 + x 3 d x4 sin tdt
Find
dx 1
What is wrong with the following calculation?
3
−1

Essential Calculus, James STEWART ()

x −1
1
= x2 −1

3
−1

1
4
=− −1=−
3
3

INTEGRALS

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Fundamental theorem of calculus
Example

x√
Find the derivative of g (x) = 0 3 + x 3 d x4 sin tdt
Find
dx 1
What is wrong with the following calculation?
3
−1

x −1
1
= x2 −1

3
−1

1
4
=− −1=−
3
3

Hint: ii. Use the Chain rule in conjunction with the fundamental theorem. iii. The Fundamental Theorem of Calculus applies to continuous functions and b

f (x)dx ≥ 0 when f ≥ 0 a Essential Calculus, James STEWART ()

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Fundamental theorem of calculus
Average value of function

With the partition as in the definition of integrals, we have the average value of f on [a, b] is
∗
∗ f (x1 ) + · · · + f (xn )
1
= lim n→∞ n→∞ b − a n n

f (xi∗ )∆x

fave = lim

1
=
b−a

i=1

b

f (x)dx a Essential Calculus, James STEWART ()

INTEGRALS

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Fundamental theorem of calculus
Average value of function

With the partition as in the definition of integrals, we have the average value of f on [a, b] is
∗
∗ f (x1 ) + · · · + f (xn )
1
= lim n→∞ n→∞ b − a n n

f (xi∗ )∆x

fave = lim

1
=
b−a

i=1

b

f (x)dx a Question: Is there a number c such that f (c) = fave ?

Essential Calculus, James STEWART ()

INTEGRALS

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Fundamental theorem of calculus
Average value of function

With the partition as in the definition of integrals, we have the average value of f on [a, b] is
∗
∗ f (x1 ) + · · · + f (xn )
1
= lim n→∞ n→∞ b − a n n

f (xi∗ )∆x

fave = lim

1
=
b−a

i=1

b

f (x)dx a Question: Is there a number c such that f (c) = fave ?

Theorem (The mean value theorem for integrals)
If f is continuous on [a, b] then there exists a number c in [a, b] such that
1
b b that is, f (c) = fave = a f (x)dx, a f (x)dx = f (c)(b − a) b−a Essential Calculus, James STEWART ()

INTEGRALS

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Fundamental theorem of calculus
Average value of function

Question: How to find the mean value c?

Essential Calculus, James STEWART ()

INTEGRALS

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Fundamental theorem of calculus
Average value of function

Question: How to find the mean value c?

Example
Find the mean value of f (x) = 1 + x 2 on [−1, 2].

Essential Calculus, James STEWART ()

INTEGRALS

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Fundamental theorem of calculus
Average value of function

Question: How to find the mean value c?

Example
Find the mean value of f (x) = 1 + x 2 on [−1, 2].
Hint: i. Compute fave : fave =

Essential Calculus, James STEWART ()

1 b−a b

f (x)dx = 2 a INTEGRALS

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Fundamental theorem of calculus
Average value of function

Question: How to find the mean value c?

Example
Find the mean value of f (x) = 1 + x 2 on [−1, 2].
Hint: i. Compute fave : fave =

1 b−a b

f (x)dx = 2 a ii. f (c) = fave so 1 + x 2 = 2.

Essential Calculus, James STEWART ()

INTEGRALS

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Fundamental theorem of calculus
Average value of function

Question: How to find the mean value c?

Example
Find the mean value of f (x) = 1 + x 2 on [−1, 2].
Hint: i. Compute fave : fave =

1 b−a b

f (x)dx = 2 a ii. f (c) = fave so 1 + x 2 = 2. iii. Find c.

Essential Calculus, James STEWART ()

INTEGRALS

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The substitution rule

Theorem (The substitution rule for definite function)
If g (x) is continuous on [a, b] and f is continuous on the range of u = g (x), then b g (b)

f (g (x))g (x)dx = a Essential Calculus, James STEWART ()

f (u)du g (a)

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The substitution rule
Symmetry

Theorem (integrals of symmetric functions)
Suppose f is continuous on [−a, a], a a
−a f (x)dx = 2 0 f (x)dx a then −a f (x)dx = 0

If f (−x) = f (x) then
If f (−x) = −f (x)

Essential Calculus, James STEWART ()

INTEGRALS

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The substitution rule
Symmetry

Theorem (integrals of symmetric functions)
Suppose f is continuous on [−a, a], a a
−a f (x)dx = 2 0 f (x)dx a then −a f (x)dx = 0

If f (−x) = f (x) then
If f (−x) = −f (x)

Essential Calculus, James STEWART ()

INTEGRALS

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Homework

Section 4.1:
Section 4.2:
Section 4.3:
Section 4.4:
Section 4.5:
Read chapter

2, 7, 10, 15,
1, 15, 17, 33, 51
13, 25, 43, 59.
7, 9, 17
13, 24, 41, 43, 57
6 to prepare for next lecture.

Essential Calculus, James STEWART ()

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