AP Problems Involving the Fundamental Theorem of Calculus

AP Problems Involving the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus:

If ,then f x a

b dx F b F a .

ddx

f t dt f x a

x .

1.

2.

)()(' xfxF

One of the hardest calculus topics to teach in the old days was Riemann sums.

They were hard to draw, hard to compute, and (many felt) totally unnecessary.

Then along came the TI graphing calculators. Using the integral utility in the CALC menu, students could actually see an integral accumulating value from left to right along the x-axis, just as a limit of Riemann sums would do:

So now we can do all kinds of summing problems before we even mention an antiderivative.

Historically, that’s what scientists had to do before calculus.

Here’s why it mattered to them:

mi/hr

hr1 4

40

20

60v(t) = 40

d = 120 mi

mi/hr

hr1 4

40

20

60

v(t)

d = 120 mi

The calculus pioneers knew that the area would still yield distance, but what was the connection to tangent lines?

And was there an easy way to find these irregularly-shaped areas?

Since the time of Archimedes, scientists had been finding areas of irregularly-shaped regions by dividing them into regularly-shaped regions. That is what Riemann sums are all about.

2.033281 2.008248 2.000329

With graphing calculators, students can find these sums without the tedium. They can also imagine the tedium of doing these sums by hand!

Best of all, they can actually see the limiting case:

And the calculator shows the thin rectangles accumulating from left to right – ideal for understanding the FTC!

a b

y=f(t)

Let us consider a positive continuous function f defined on [a, b].

Choose an arbitrary x in [a, b].

x

Each choice of x determines a unique area from a to x, denoted as usual by

( )x

af t dt

a b

y=f(t)

x

So ( ) ( ).x

a

d f t dt f xdx

But that is only half the story.

Now that we know that is an antiderivative of f,we know that it differs from any antiderivative of f by a constant.

That is, if F is any antiderivative of f,

( )x

af t dt

( ) ( ) .x

af t dt F x C

To find C, we can plug in a:

( ) ( )

0 ( )( )

a

af t dt F a C

F a CC F a

So ( ) ( ) ( ).x

af t dt F x F a

Now plug in b:

( ) ( ) ( ).b

af t dt F b F a

This was the FTC. This was the result that changed the world.

2.000329

Now, instead of wasting a full afternoon just to get an approximation of the area under one arch of the sine curve, you could find one antiderivative, plug in two numbers, and subtract!

0sin( ) cos( ) cos( ) cos(0) 2.

0x dx x

Since 2000, the AP Calculus Test Development Committee has been emphasizing a conceptual understanding of the definite integral, resulting in these “new” problem types:

Functions defined as integrals

Accumulation Problems

Integrals from Tables

Finding , given and

Interpreting the Definite Integral( )f b ( )f a ( )f x

Problem of the:

Suppose 3

1( ) 2

xf t dt x x k .

(a) Find f(x). (b) Find k.

3

1

2

( ) 2 )

( ) 3 2

xd df t dt x x kdx dx

f x x

(a) By the Fundamental Theorem,

(b) Plug in x = 1:1 3

1( ) 1 2(1)

0 11

f t dt k

kk

If 2( )f x x , which of the following could be the graph of

1( )

xy f t dt ?

(A)

(B)

(C)

(D)

(E)

Here was the problem (1987):

This problem had been checked:

1. by the author who had written it;

2. by the committee that had okayed it;

3. by the committee that had okayed it for a pre-test;

4. by the ETS test development specialists;

5. The committee, reviewing the final form of the college pre-test.

The proposed key was (B). That is,

While everyone was concentrating on the Fundamental Theorem application, they had missed the hidden “initial condition” that y must equal zero when x = 1!

If 2( )f x x , the graph of 1

( )x

y f t dt could be

(B)

Here’s 1995 / BC-6:

12

-1-2-3

1 2 3 4 50

Let f be a function whose domain is the closed interval [0, 5]. The graph of f is shown above.

Let 3

20

( ) ( ) .x

h x f t dt

(a) Find the domain of h. (b) Find (2).h (c) At what x is h(x) a minimum? Show the analysis that leads to your conclusion.

(a) The domain of h is all x for which

is defined:32

0( )

x

f t dt

0 3 5 6 4

2x x

(b) A little Chain Rule:

32

0

1( ) ( ) 32 2

1 3(2) (4)2 2

xd xh x f t dt fdx

h f

12

-1-2-3

1 2 3 4 50

Let f be a function whose domain is the closed interval [0, 5]. The graph of f is shown above.

Let 3

20

( ) ( ) .x

h x f t dt

(a) Find the domain of h. (b) Find (2).h (c) At what x is h(x) a minimum? Show the analysis that leads to your conclusion.

(c) Since is positive from

-6 to -1 and negative from -1 to 4, the minimum occurs at an endpoint. By comparing areas, h(4) < h(-6) = 0, so the minimum occurs at x = 4.

32

0( )

x

f t dt

This “area comparison” genre of problem was pretty common in the early graphing calculator days.

The amount of water in a storage tank, in gallons, is modeled by a continuous function on the time interval 0 7t , where t is measured in hours. In this model, rates are given as follows: (i) The rate at which water enters the tank is 2( ) 100 sinf t t t

gallons per hour for 0 7t . (ii) The rate at which water leaves the tank is

250 for 0 3

( ) gallons per hour.2000 for 3 7

tg t

t

The graphs of f and g, which intersect at t = 1.617 and t = 5.076, are shwn in the figure above. At time t = 0, the amount of water in the tank is 5000 gallons.

500

2000

1000

1500

2500

1 2 3 4 5 6 7

2007 / AB-2 BC-2

(a) How many gallons of water enter the tank during the time interval 0 7t ? Round your answer to the nearest gallon. (b) For 0 7t , find the time intervals during which the amount of water in the tank is decreasing. Give a reason for each answer. (c) For 0 7t , at what time t is the amount of water in the tank greatest? To the nearest gallon, compute the amount of water at this time. Justify your answer.

500

2000

1000

1500

2500

1 2 3 4 5 6 7

500

1 2 3 4 5 6 7

1000

1500

2000

2500

The standard description of the FTC is that

“The two central operations of calculus, differentiation and integration, are inverses of each other.” —Wikipedia

A more useful description is that the two definitions of the definite integral:•The difference of the values of an anti-derivative taken at the endpoints, [definition used by Granville (1941) and earlier authors]•The limit of a Riemann sum, [definition used by Courant (1931) and later authors]

yield the same value.

2004 AB3(d)A particle moves along the y-axis so that its velocity v at time t ≥ 0 is given by v(t) = 1 – tan–1(et). At time t = 0, the particle is at y = –1. Find the position of the particle at time t = 2.

y '(t) = v(t) = 1 – tan–1(et)

y(t) = ?

Velocity Time = Distance

time

velo

city

dist

a nce

The Fundamental Theorem of Calculus (part 1):

If ,then f x a

b dx F b F a .)()(' xfxF

Change in y-value equals

Since we know that y(0) = –1:

v t dt 1 tan 1 e t 0

2

0

2

dt 0.3607,

y 2 y 0 0.3607 1.3607

If we know an anti-derivative, we can use it to find the

value of the definite integral.

All students should know how to interpret the following applications as accumulations:

Areas (sums of rectangles)

Volumes (sums of regular-shaped slices)

Displacements (sums of v(t)∙∆t)

Average values (Integrals/intervals)

BC: Arclengths (sums of hypotenuses)

BC: Polar areas (sums of sectors)

Problem of the Day :

The population density of the city of Washerton decreases as you move away from the city center. In fact, it can be approximated (in people per square mile) by the function 10,000(2 – r) at a distance r miles from the city center. (a) What is the radius of the populated portion of the city? (b) A thin ring around the center of the city has thickness r and radius r. What is its area? [Hint: Imagine straightening it out to make a thin rectangular strip.] (c) What is the population of the strip in part (b)? (d) Estimate the total population of Washerton by setting up and evaluating a definite integral.

The population density of the city of Washerton decreases as you move away from the city center. In fact, it can be approximated (in people per square mile) by the function 10,000(2 – r) at a distance r miles from the city center. (a) What is the radius of the populated portion of the city?

(a) r = 2 miles.

(b) A thin ring around the center of the city has thickness r and radius r. What is its area? [Hint: Imagine straightening it out to make a thin rectangular strip.]

(b) A = 2πr Δr

(d) Estimate the total population of Washerton by setting up and evaluating a definite integral.

people per sq. mile sq. mil

2

0es

10,000(2 )(2 )

83,776 people

r r dr

(d)

Another implication of the Fundamental Theorem (and a source of several recent problems that have caused trouble for students):

( ) ( ) (

( ) ( )

)

( )b

a

b

af x dx f b f a

f b f a f x dx

Thus, given f(a) and the rate of change of f on [a, b], you can find f(b).

The Kicker in 2003 / AB-4 BC-4:

(4, –2)

(–3, 1)

–4 –2 2 4

–2

2

x

y

x

Let f be a function defined on the closed interval 3 4x with f(0) = 3. The graph of f , the derivative of f, consists of one line segment and a semicircle, as shown above. (d) Find f(–3) and f(4). Show the work that leads to your answers.

3

0

4

0

( 3) (0) ( )

1 13 ( 2)( 2) ( 1)(1)2 2

92

(4) (0) ( )

3 ( 8) ( 2 )2 5

f f f x dx

f f f x dx

Problem of the Day #35:

If 2( ) sin ( )F x x and F(2) = 5, find F(7).

Solve this just like the last one was solved.