MAT137 j Calculus! j Lecture 28 - math.toronto.edu. FT… · FTC j \Area so Far" function The fundamental theorem of calculus deals with functions of the form g(x)=S x a f(t)dt; where

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MAT137 | Calculus! | Lecture 28

Today:

AntiderivativesFundamental Theorem of Calculus

Next:

More FTC(review v. 8.5 - 8.7)§5.7 Substitution(v. 9.1 - 9.4)

official website http://uoft.me/MAT137Beatriz Navarro-Lameda L0601 MAT137 16 January 2018

http://uoft.me/MAT137

Properties of the Definite Integral

Let f and g be integrable functions, and let a,b, c be any real numbers.

1 [order of limits] ∫

b

af (x) dx = −∫

a

bf (x) dx

a b

+a b

−

2 [constant multiple] ∫

b

acf (x) dx = c ∫

b

af (x) dx

a ba b

y = cf (x)

y = f (x)

a ba b

y = cf (x)

y = f (x)

Beatriz Navarro-Lameda L0601 MAT137 16 January 2018

3 [sum] ∫

b

a(f (x) + g(x)) dx = ∫

b

af (x) dx + ∫

b

ag(x) dx

a b

y = f (x)

+

a b

y = g(x)

=

a ba b

y = f (x) + g(x)

4 [additivity] ∫

b

af (x) dx = ∫

c

af (x) dx + ∫

b

cf (x) dx

a bca b

y = f (x)

∫c

af (x)dx ∫

b

cf (x)dx


Comparison Properties of the Integral

The following properties are true only if a ≤ b.

5 [integral of a non-negative function]

f (x) ≥ 0 on [a,b] Ô⇒ ∫

b

af (x) dx ≥ 0.

6 [domination]

If f (x) ≥ g(x) on [a,b] Ô⇒ ∫

b

af (x) dx ≥ ∫

b

ag(x) dx .



Example 1

If

∫

1

−1f (x)dx = 2, ∫

5

1f (x)dx = 3, ∫

1

−1g(x)dx = 5, ∫

0

−1g(x)dx = 1,

find each of the following integrals, if possible:

(a) ∫1

−1(2f (x) + g(x)) dx

(b) ∫5

−1f (x) dx

(c) ∫1

0f (x) dx

(d) ∫1

0g(x) dx



If f is continuous and f (x) < 0 for all x ∈ [a,b], then ∫ba f (x)dx

1 must be negative

2 might be 0

3 not enough information



Let f be a continuous function on the interval [a,b].

True or False.

There exist two constants m and M, such thatm(b − a) ≤ ∫

ba f (x)dx ≤M(b − a)


Antiderivatives

Definition (Antiderivative)

A function F is an antiderivative of a function f on an interval I if

F ′(x) = f (x) for all x in I .


Antiderivatives

Theorem (General form of antiderivative)

If F is an antiderivative of f on an interval I , then the most generalantiderivative of f on I is

∫ f (x)dx = F (x) + C

where C ∈ R is an arbitrary constant.

Note: ∫ f (x)dx represents the collection of all functions whose

derivative is f (x).

Example 2

Find a function f such that f ′(x) = 3x2 and f (0) = 1.


Antiderivatives

Example 3

Find the antiderivative of f (x) = (3x + 5)7.

Here’s the general strategy in the form of a flow diagram:

guess checkclose

not close

adjust checkcorrect

not quite correct

writemost

generalantiderivative


Antiderivatives

Example 4

Find a function f (x) if f ′′(x) = sin x + ex − x2, and f (0) = 0, f ′(0) = 2.


Indefinite Integral - Guess and Check

Example 5

Evaluate

∫ex

ex + 1dx



Example 6

Evaluate

∫sin

√x

√x

dx



Example 7

Evaluate

∫ sin(x) cos(x)dx


The fundamental theorem of calculus deals with functions of the form

g(x) = ∫x

af (t) dt,

where f is a continuous function on [a,b] and x varies between a and b.

For example, if f is non-negative, then g(x) can be interpreted as the areaunder the graph of f between a and x , where x varies from a to b. Youcan think of g as “the area so far” function.


Area so Far Function

Let f (t) = t and a = 0, then the function

g(x) = ∫x

0tdt

represents the area under the curve in the picture. Thus,

g(x)

x



Below is the graph of a function f .

-1 -0.5 0.5 1 1.5 2 2.5

1

2

3

4

Let g(x) = ∫x

0f (t)dt. Then for 0 < x < 2, g(x) is

1 increasing and concave up.

2 increasing and concave down.

3 decreasing and concave up.

4 decreasing and concave down.



Below is the graph of a function f .

-1 -0.5 0.5 1 1.5 2 2.5

1

2

3

4

Let g(x) = ∫x

0f (t)dt. Then

1 g(0) = 0, g ′(0) = 0 and g ′(2) = 0

2 g(0) = 0, g ′(0) = 4 and g ′(2) = 0

3 g(0) = 1, g ′(0) = 0 and g ′(2) = 1

4 g(0) = 0, g ′(0) = 0 and g ′(2) = 1


FTC | “Area so Far” function

The fundamental theorem of calculus deals with functions of the form

g(x) = ∫x

af (t) dt,

where f is a continuous function on [a,b] and x varies between a and b.Remark: t is a ”dummy variable”

For example, if f is non-negative, then g(x) can be interpreted as the areaunder the graph of f between a and x , where x varies from a to b.You can think of g as “the area so far” function.




g(x) = ∫x

0t dt

represents the area under the curve in the picture.

g(x)

x

g(x) = ∫x

0t dt = 1

2x ⋅ x =1

2x2.

What is g ′(x)?




g(x) = ∫x

0t dt


g(x)

x

g(x) = ∫x

0t dt = 1

2x ⋅ x =1

2x2.

What is g ′(x)?




g(x) = ∫x

0t dt


g(x)

x

g(x) = ∫x

0t dt = 1

2x ⋅ x =1

2x2.

What is g ′(x)?Beatriz Navarro-Lameda L0601 MAT137 16 January 2018

FTC | Part I

Theorem (Fundamental Theorem of Calculus, Part 1 [FTC1])

If f is continuous on an interval [a,b], then the function g defined by

g(x) = ∫x

af (t) dt for a ≤ x ≤ b

is continuous on [a,b] and differentiable on (a,b).Moreover, g ′(x) = f (x).

Roughly speaking, this says the following: when f is continuous, if we firstintegrate and then differentiate, we get f back.


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MAT137 j Calculus! j Lecture 28 - math.toronto.edu. FT… · FTC j \Area so Far" function The fundamental theorem of calculus deals with functions of the form g(x)=S x a f(t)dt; where