Math 221 Week 11 part 1

Fundamental Theorem of Calculus: intuition

Please take a moment to just breathe.

In this section we investigate the “2nd” part of the Fundamental Theorem of Calculus.

Recall the definition:

The definite integral of � from � to � is

�

if this limit exists.

If the limit exists, we say that � is integrable on � .

We can think of the integral as the signed area under the curve � between � and � .

f(x) x = a x = b

∫b

af(x)dx = lim

n→∞

n

∑i=1

f(x*i )Δx

f [a, b]

y = f(x) x = a x = b

Suppose that you are flying due west in a small airplane, and your GPS is broken. You have to calculate your location by hand, as pilots did before 1973.

Every 10 minutes, you measure your airspeed in kilometers per minute. (Assume that the radio says there is no wind.)

Suppose you know that your airspeeds are as follows:

0 3.3 kilometers/min

10 3.2

20 3.1

30 3.0

40 2.8

50 2.7

How far have you traveled? time

airspeed

time airspeed

0 3.3 kilometers per minute

10 3.2

20 3.1

30 3.0

40 2.8

50 2.7

In the first 10 minutes,

you go approximately

33 kilometers.

The area under the velocity

graph from � to �

is also approximately 33 units.

In both cases, we’re taking the product (10 min)(3.3 kpm)

t = 0 t = 10time

airspeed

time airspeed

0 3.3 kilometers per minute

10 3.2

20 3.1

30 3.0

40 2.8

50 2.7

In the second 10 minutes,

you go approximately

32 kilometers.

The area under the velocity

graph from � to �

is also approximately 32 units.

In both cases, we’re taking the product (10 min)(3.2 kpm)

t = 10 t = 20time

airspeed

time airspeed

0 3.3 kilometers per minute

10 3.2

20 3.1

30 3.0

40 2.8

50 2.7

In the last 10 minutes,

you go approximately

28 kilometers.

The area under the velocity

graph from � to �

is also approximately 28 units.

In both cases, we’re taking the product (10 min)(2.8 kpm)

t = 50 t = 60time

airspeed

Adding these six distances up, we get

3.3(10) + 3.2(10) + 3.1(10) + 3.0(10) + 2.8(10) + 2.7(10) = 181

So the total distance traveled, and the total area under the graph, is about 181 km.

More generally, if we know the velocity function � , we can approximate the distance travelled from time � to � by using � subintervals as follows:

let �

Then the total distance travelled is approximately

�

This expression is also a Riemann sum for the area under � between � and � .

v(t)t = a

t = b n

Δt =b − a

n

n

∑i=1

v(a + iΔt)Δt

y = v(t) t = a t = b

If we let the length of the time subintervals go to zero by letting � go to infinity, then the distance becomes

� �

So we get the formula

�

where � is the position at time � .

“The net change in position is equal to the definite integral of the velocity.”

n

limn→∞

n

∑i=1

v(a + iΔt)Δt = ∫b

av(t)dt

∫b

av(t)dt = p(b) − p(a)

p(t) t

Fundamental Theorem of Calculus (part 2)

If � is continuous on � , then

�

where � is any antiderivative of � .

v [a, b]

∫b

av(t)dt = p(b) − p(a)

p(t) v(t)

Fundamental Theorem of Calculus (part 2) using the book’s letters:

If � is continuous on � , then

�

where � is any antiderivative of � .

f [a, b]

∫b

af(t)dt = F(b) − F(a)

F f

Example: Find �

We know that one antiderivative of �

is � .

So by the Fundamental Theorem of Calculus,

�

�

∫1

0x2dx

f(x) = x2

F(x) =13

x3

∫1

0x2dx = F(1) − F(0)

=13

(1)3 −13

(0)3 =13

Example: Find �

What happens if we use a different antiderivative? Eg:

�

By the Fundamental Theorem of Calculus,

�

�

∫1

0x2dx

F(x) =13

x3 + 1000

∫1

0x2dx = F(1) − F(0)

= ( 13 (1)3 + 1000) − ( 13 (0)3 + 1000) = 13

Example: Find �

Try this one yourself (pause the video)

∫3

2(x4 +

1x

) dx

Example: Find �

One antiderivative of � is

�

So �

�

∫3

2(x4 +

1x

) dx

x4 +1x

F(x) =x5

5+ ln(x)

∫3

2(x4 +

1x

) dx = F(3) − F(2)

= ( 35

5+ ln(3)) − ( 2

5

5+ ln(2)) = 2115 + ln(3/2)