Section 5.3Evaluating Definite Integrals

Math 1a

December 10, 2007

Announcements

I my next office hours: Monday 1–2, Tuesday 3–4 (SC 323)

I MT II is graded. Come to OH to talk about it

I Final seview sessions: Wed 1/9 and Thu 1/10 in Hall D, Sun1/13 in Hall C, all 7–8:30pm

I Final tentatively scheduled for January 17, 9:15am

Outline

FTC2ProofExamples

Total Change

Indefinite IntegralsMy first table of integrals

Examples“Negative Area”

Theorem (The Second Fundamental Theorem of Calculus,Strong Form)

Suppose f is integrable on [a, b] and f = F ′ for another function f ,then ∫ b

af (x) dx = F (b)− F (a).

Proof.We will choose Riemann sums which converge to the right-handside. Let n be given. On the interval [xi−1, xi ] there is a point ci

such that

f (ci ) = F ′(ci ) =F (xi )− F (xi−1)

xi − xi−1

Then

Rn =n∑

i=1

f (ci )∆x =n∑

i=1

(F (xi )− F (xi−1)

)= F (xn)− F (x0) = F (b)− F (a).

So Rn → F (b)− F (a) as n→∞.

Examples

Find the following integrals:

I

∫ 1

0x2 dx ,

∫ 1

0x3 dx ,

∫ 1

0xn dx (n 6= −1),

∫ 2

1

1

xdx

I

∫ π

0sin θ dθ,

∫ 1

0ex dx ,

∫ π/4

0tan θ dθ

Outline

FTC2ProofExamples

Total Change

Indefinite IntegralsMy first table of integrals

Examples“Negative Area”

The Integral as Total Change

Another way to state this theorem is:∫ b

aF ′(x) dx = F (b)− F (a),

or the integral of a derivative along an interval is the total changebetween the sides of that interval. This has many ramifications:

The Integral as Total Change

Another way to state this theorem is:∫ b

aF ′(x) dx = F (b)− F (a),

or the integral of a derivative along an interval is the total changebetween the sides of that interval. This has many ramifications:

TheoremIf v(t) represents the velocity of a particle moving rectilinearly,then ∫ t1

t0

v(t) dt = s(t1)− s(t0).

The Integral as Total Change

Another way to state this theorem is:∫ b

aF ′(x) dx = F (b)− F (a),

or the integral of a derivative along an interval is the total changebetween the sides of that interval. This has many ramifications:

TheoremIf MC (x) represents the marginal cost of making x units of aproduct, then

C (x) = C (0) +

∫ x

0MC (q) dq.

The Integral as Total Change

Another way to state this theorem is:∫ b

aF ′(x) dx = F (b)− F (a),

or the integral of a derivative along an interval is the total changebetween the sides of that interval. This has many ramifications:

TheoremIf ρ(x) represents the density of a thin rod at a distance of x fromits end, then the mass of the rod up to x is

m(x) =

∫ x

0ρ(s) ds.

Example

If oil leaks from a tank at a rate of r(t) gallons per minute at time

t, what does

∫ 120

0r(t) dt represent?

SolutionThe amount of oil lost in two hours.

Example

If oil leaks from a tank at a rate of r(t) gallons per minute at time

t, what does

∫ 120

0r(t) dt represent?

SolutionThe amount of oil lost in two hours.

Outline

FTC2ProofExamples

Total Change

Indefinite IntegralsMy first table of integrals

Examples“Negative Area”

A new notation for antiderivatives

To emphasize the relationship between antidifferentiation andintegration, we use the indefinite integral notation∫

f (x) dx

for any function whose derivative is f (x).

Thus∫x2 dx = 1

3x3 + C .

A new notation for antiderivatives

To emphasize the relationship between antidifferentiation andintegration, we use the indefinite integral notation∫

f (x) dx

for any function whose derivative is f (x). Thus∫x2 dx = 1

3x3 + C .

My first table of integrals∫[f (x) + g(x)] dx =

∫f (x) dx +

∫g(x) dx∫

xn dx =xn+1

n + 1+ C (n 6= −1)∫

ex dx = ex + C∫sin x dx = − cos x + C∫cos x dx = sin x + C∫

sec2 x dx = tan x + C∫sec x tan x dx = sec x + C∫

1

1 + x2dx = arctan x + C

∫cf (x) dx = c

∫f (x) dx∫

1

xdx = ln x + C∫

ax dx =ax

ln a+ C∫

csc2 x dx = − cot x + C∫csc x cot x dx = − csc x + C∫

1√1− x2

dx = arcsin x + C

Outline

FTC2ProofExamples

Total Change

Indefinite IntegralsMy first table of integrals

Examples“Negative Area”

Example

Find the area between the graph of y = (x − 1)(x − 2), the x-axis,and the vertical lines x = 0 and x = 3.

Solution

Consider

∫ 3

0(x − 1)(x − 2) dx. Notice the integrand is positive on

[0, 1) and (2, 3], and negative on (1, 2). If we want the area of theregion, we have to do

A =

∫ 1

0(x − 1)(x − 2) dx −

∫ 2

1(x − 1)(x − 2) dx +

∫ 3

2(x − 1)(x − 2) dx

=[

13x3 − 3

2x2 + 2x]10−[

13x3 − 3

2x2 + 2x]21

+[

13x3 − 3

2x2 + 2x]32

=5

6−(−1

6

)+

5

6=

11

6.

Example

Find the area between the graph of y = (x − 1)(x − 2), the x-axis,and the vertical lines x = 0 and x = 3.

Solution

Consider

∫ 3

0(x − 1)(x − 2) dx. Notice the integrand is positive on

[0, 1) and (2, 3], and negative on (1, 2). If we want the area of theregion, we have to do

A =

∫ 1

0(x − 1)(x − 2) dx −

∫ 2

1(x − 1)(x − 2) dx +

∫ 3

2(x − 1)(x − 2) dx

=[

13x3 − 3

2x2 + 2x]10−[

13x3 − 3

2x2 + 2x]21

+[

13x3 − 3

2x2 + 2x]32

=5

6−(−1

6

)+

5

6=

11

6.