The Fundamental Theorem of Calculus
By:
Matthew Gruda
Vinay Ramesh
1
Table of Contents
First and Second Fundamental Theorems of Calculus……………………3
Analytical Examples…………………………………………………………………………………………7
AP Level Multiple Choice………………………………………………………………………………9
Conceptual Example………………………………………………………………………………………11
Real World Applicability………………………………………………………………………………….12
Graph Example……………………………………………………………………………………………………13
AP Level FRQ………………………………………………………………………………………………………15
Who discovered this………………………………………………………………………………………….18
2
First FTC Explanation:
The first fundamental
theorem says that the area under
the curve from 𝑎 to 𝑏 of the
function 𝑓(𝑥) with the antiderivative 𝐹(𝑥) equals 𝐹(𝑎) −
𝑏
𝐹(𝑏). Thus ∫𝑎 𝑓(𝑥)𝑑𝑥 = 𝐹(𝑏) −
𝐹(𝑎).
Proof of the Second Fundamental Theorem of Calculus
So you know that to solve Second Fundamental Theorem problems, when asked to
take the derivative, you simply plug in the upper bound of the integral into the
function and then multiply that equation by the derivative of the upper bound. But
why is that the case?? Here’s why:
y = f(t)
𝑎
𝑥
𝑥 + ∆𝑥
𝑏
3
𝑥
Let 𝐹(𝑥) = ∫𝑎 𝑓(𝑡)𝑑𝑡 where 𝑎 ≤ 𝑥 ≤ 𝑏
Now, we can differentiate using the limit definition of derivatives.
Eq.1
𝑥+∆𝑥
∫𝑎
𝐹 ′ (𝑥) = lim
𝑥
𝑓(𝑡)𝑑𝑡 − ∫𝑎 𝑓(𝑡)𝑑𝑡
∆𝑥
∆𝑥→0
The numerator can be simplified into one integral, because both terms in the
numerator share the same lower bound. A new integral expression combining the
two of these terms can be written with the lower bound equaling 𝑥 and the upper
bound equaling 𝑥 + ∆𝑥.
Eq. 2
𝑥+∆𝑥
𝐹
′ (𝑥)
= lim
∫𝑥
𝑓(𝑡)𝑑𝑡
∆𝑥
∆𝑥→0
1 𝑥+∆𝑥
= lim
∫
𝑓(𝑡)𝑑𝑡
∆𝑥→0 ∆𝑥 𝑥
This expression is extremely similar to definition of the Mean Value Theorem. The
Mean Value Theorem states that if you have a continuous and differentiable
function in an interval, in this case [𝑥, 𝑥 + ∆𝑥] there exists some value c in this
𝑥+∆𝑥
interval such that 𝑓(𝑐)∆𝑥 = ∫𝑥
𝑓(𝑡)𝑑𝑡. In other words, there is some rectangle
with some height 𝑓(𝑐), and width ∆𝑥, whose area is equal to the area under the
curve within the interval [𝑥, 𝑥 + ∆𝑥].
Eq. 3
We can rewrite this expression as 𝑓(𝑐) =
1
𝑥+∆𝑥
∫
∆𝑥 𝑥
𝑓(𝑡)𝑑𝑡
Look familiar? Good, because we can plug in 𝑓(𝑐) into Equation 2 above.
Eq. 4
𝐹 ′ (𝑥) = lim 𝑓(𝑐)
∆𝑥→0
We are at the home stretch!
4
Now, it is important to note that 𝑐 → 𝑥 𝑎𝑠 ∆𝑥 → 0 because the boundary for which
c can lie in becomes smaller and smaller, eventually leading up to 𝑐 = 𝑥.
Therefore, 𝑓(𝑐) → 𝑓(𝑥) 𝑎𝑠 ∆𝑥 → 0.
With this information, we can see that 𝐹 ′ (𝑥) = 𝑓(𝑥)
This is HUGGEE!!! This tells us that an integral of a derivative of a function is
equal to the function itself. This fact ties together differential and integral calculus.
ONE MORE THING TO NOTE!!!!!!
When solving 2nd Fundamental Theorem of Calculus problems, make sure you
multiply the answer by the derivative of the upper bound.
This is why:
Using the First Fundamental Theorem of Calculus, we are familiar with the
following process:
Eq. 5
𝑓(𝑥)
∫
𝑔′ (𝑡)𝑑𝑡 = 𝑔(𝑓(𝑥)) − 𝑔(𝑎)
𝑎
Now, when we differentiate this equation, we see why simply plugging in the
upper bound is not enough
Eq. 6
𝑓(𝑥)
𝑑
𝑑
[∫
𝑔′ (𝑡)𝑑𝑡] =
[𝑔(𝑓(𝑥)) − 𝑔(𝑎)]
𝑑𝑥 𝑎
𝑑𝑥
Therefore,
5
𝑓(𝑥)
𝑑
[∫ 𝑔′ (𝑡)𝑑𝑡] = 𝑔′ (𝑓(𝑥)) ∗ 𝑓 ′ (𝑥) − 0
𝑑𝑥 𝑎
AHAAA!! That 𝑓 ′ (𝑥) tell us that after we plug in the upper bound for t, we must
multiply the equation by the derivative of the upper bound as well.
Now you know why the Second Fundamental Theorem works. YOU’RE
READY!!!!
Now let’s move on to some examples.
6
Analytical Examples
𝑡2
1. Find f ’(t) if f(t) = ∫3𝜋 √𝑥 + 1 𝑑𝑥
First, take the derivative of f(x)
f’(t) =
𝑡2
𝑑
𝑑𝑥
[∫3𝜋 √𝑥 + 1 𝑑𝑥]
Plug in the upper limit for “x”
f’(t) = √𝑡 2 + 1
BUT WAIT!!!!!
Don’t forget to multiply the equation by the derivative of the upper bound!
f’(t) = √𝑡 2 + 1 * 2t
You got it! This is the Second Fundamental Theorem of Calculus.
𝜋/3
2. Find the value of ∫𝜋/6 (2 sin(3𝑥)) 𝑑𝑥
For simplicity, take the “2” out of the integral, as it is a constant
𝜋/3
2∫𝜋/6 (sin(3𝑥)) 𝑑𝑥
Use u substitution to help solve this integral. In this case, let 𝑢 = 3𝑥
1
Therefore, 𝑑𝑢 = 3𝑑𝑥, and 𝑑𝑢 = 𝑑𝑥
3
Now, create new u-limits:
𝜋
Upper bound: 𝑢 = 3 ( ) ; Therefore, upper bound is u=𝜋
3
𝜋
𝜋
Lower bound: 𝑢 = 3 ( ) ; Therefore, lower bound is u=
6
2
Rewrite the integral with the u substitution
1
𝜋
2 ∗ ∫𝜋/2 sin(𝑢) 𝑑𝑢
3
7
Integrate sin(u):
The integral of 𝑠𝑖𝑛(𝑢) = − 𝑐𝑜𝑠(𝑢)
Rewrite the equation and use the First Fundamental Theorem of Calculus
2
(− cos(𝑢) |𝜋𝜋/2
3
2
𝜋
((−𝑐𝑜𝑠(𝜋)) − (− cos ( ))
3
2
− cos(𝜋) = 1
𝜋
− cos ( ) = 0
2
2
2
3
3
The equation now becomes (1 − 0) 𝑜𝑟
𝜋/3
Therefore, ∫𝜋/6 (2 sin(3𝑥)) 𝑑𝑥 =
2
3
8
AP Level Multiple Choice:
3𝑥
1) Find 𝑓 ′ (𝑥) if 𝑓(𝑥) = ∫2𝜋 (𝑡 4 + 4)𝑑𝑡
a. 𝑓 ′ (𝑥) = 𝑥 4 + 4
b. 𝑓 ′ (𝑥) = 3𝑥 4 + 4
c. 𝑓 ′ (𝑥) = 3𝑥 4 + 12
d. 𝑓 ′ (𝑥) = 4𝑥 3
e. 𝑓 ′ (𝑥) = 3(3𝑥)4 + 12
The correct answer is C.
Explanation:
𝑑𝑦
𝑑𝑥
[𝑓(𝑥)] =
𝑑𝑦
𝑑𝑥
3𝑥
[∫2𝜋 (𝑡 4 + 4)𝑑𝑡]
Differentiate 𝑓(𝑥)
𝑓 ′ (𝑥) = [(3𝑥)3 + 4] + 4 ∗ (3) Substitute 3𝑥 for 𝑡 and multiply
by the derivative of 3x, 3
𝑓 ′ (𝑥) = 3(3𝑥)4 + 12
2) Find the area bounded by 𝑦 = 0, 𝑥 = 1, 𝑥 =
Simplify
𝜋
2
and 𝑦 = 𝑠𝑖𝑛𝑥
a. -1
b. 1
c. .540
d. -.541
e. .541
The correct answer is C.
9
First, use the First Fundamental Theorem of Calculus to turn this into
an integral. The FTC gives the area from the curve to the x-axis
between two points, so
𝜋
2
∫1 𝑠𝑖𝑛𝑥 𝑑𝑥
(−𝑐𝑜𝑠𝑥|1𝜋
⁄2
Write as definite Intergral
Integrate
(0 − (−cos(1))
Expand
.540
Simplify
10
Conceptual Example:
A bay is 10 miles wide and it’s
shore can be approximated
with the function (𝑥) =
−.1(𝑥 − 4)2 + 5 . Find the
area of the bay.
Answer: 40.66 square miles
Explanation:
(Source: http://www.sunsail.com/files/Destinations/pacific/australia/whitsundays/butterflybay-hook-island.jpg)
Guess what? Area under the curve! FTC!
First find the bounds. Looking at the graph, it goes from 0 to 10 miles.
Next, setup the integral:
10
∫0 −.1(𝑥 − 4)2 + 5 𝑑𝑥
10
∫0 −.1𝑥 2 + .8𝑥 − 1.6 + 5 𝑑𝑥
−.1𝑥 3
(
3
2
Simplify
10
+ 4𝑥 + 3.4𝑥)
Integrate
40.666
Evaluate
0
11
Now where the heck could you use this in real life?
Real World Applicability:
The First Fundamental Theorem of Calculus can be used to calculate the
area of non-regular shapes, and since non-regular shapes are quite regular in the
real world, the FTC is very useful there. The FTC also underlies half of calculus,
including useful concepts such as volumes of solids of revolution, which can
calculate the volume from a function. The FTC can also be used to calculate
displacement from velocity, and velocity from acceleration. For example, it can
be used to calculate a ship’s position from a starting point and a velocity function,
making it very useful in navigation.
12
Graph Example:
(Source: Collegeboard 2014 AP Calculus AB Exam)
The function f is defined on the closed interval [-5,4]. The graph of f consists
of three line segments and is shown in the figure above. Let g be the function
𝑥
defined by 𝑔(𝑥) = ∫0 (𝑓(𝑡))𝑑𝑡.
Find g’ (4)
In order to figure this out, you must first differentiate the above equation: 𝑔(𝑥) =
𝑥
∫0 (𝑓(𝑡))𝑑𝑡
This can be easily accomplished by using the Second Fundamental Theorem of
Calculus
𝑔
′ (𝑥)
𝑥
𝑑
=
[∫ (𝑓(𝑡))𝑑𝑡]
𝑑𝑥 0
𝑔′ (𝑥) = 𝑓(𝑥)*
𝑑
𝑑𝑥
[𝑥] = 𝑓(𝑥) ∗ 1
This realization is the most important part of the problem.
Now that we have established that 𝑔′(𝑥) = 𝑓(𝑥), 𝑔′ (4)𝑚𝑢𝑠𝑡 𝑏𝑒 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 𝑓(4).
13
On the graph, it is clear that the value of f at x=4 is -4, or 𝑓(4) = −4
Therefore, 𝑔′ (4) = −4
14
AP Level FRQ Part:
The function f is defined on the interval −4 ≤ 𝑥 ≤ 4. The graph of f includes one
𝑥
quarter circle. Let 𝑔(𝑥) = 3 ∫0 𝑓(𝑡) 𝑑𝑡 + 3𝑥
a). Find g(-3) and g'(-3).
b). Find the x-value of the absolute minimum of 𝑔(𝑥) on the interval −4 ≤ 𝑥 ≤ 4.
Answer and Explanation:
a).
−3
𝑔(−3) = 3 ∫0 𝑓(𝑡) 𝑑𝑡 + 3(−3) = 3 ∗ (1 − 𝜋) − 9 = −6 − 3𝜋
15
𝑔′ (𝑥) = 3𝑓(𝑥) + 3
𝑔′ (−3) = 3𝑓(−3) + 3 = 3 ∗ 2 + 3 = 9
Step 1: plug in -3 for 𝑥 in 𝑔(𝑥).
Step 2: Remember, the integral is the area under the curve, and thus you can
find the area under the curve geometrically. Thus use the formula for a circle
1
(𝜋𝑟 2 ) and a triangle ( 𝑏ℎ) to get the area.
2
Step 3: Simplify.
Step 4:
Use the second FTC to derive 𝑔′(𝑥). Remember how 2FTC
works: the derivative of an integral equals the inner part times the derivative of 𝑥.
Step 5: Plug in -3 to 𝑔′(𝑥) and simplify.
b).
The absolute minimum is -2.
Step 1: the absolute minimum will occur when 𝑔′ (𝑥) = 0 or at the
endpoints.
Step 2: find where 𝑔′ (𝑥) = 0. Remember what 𝑔′(𝑥) is from part a.
0 = 3𝑓(𝑥) + 3
−1 = 𝑓(𝑥)
Critical points = 1, −√3
Step 3:
−√3 can be eliminated because the graph of 𝑔′(𝑥) would go
from positive to negative there, which means −√3 is a relative maximum.
Step 4:
𝑔(−4) = 3 − 3𝜋
𝑔(1) = 6 − 3
16
𝑔(4) = 0 + 6
17
Now who the heck could come up with this crap?
Discovered By: Sir Isaac Newton
Sir Isaac Newton, born January 4, 1643 in England,
was a brilliant mathematician and physicist. He did
not want to be a farmer like his father, so got into
Trinity College in Cambridge. The university
closed in 1665 because of the plague, so he went
home and developed calculus to describe the laws
of motion, the orbits of planets, optics and the
theory of universal gravitation. He then published
his work in the Principia, which set off a
controversy with Gottfried Leibniz who clamed to
have developed calculus first. In his later years, he
became Master of the Royal Mint and president of the Royal Society, and died on
March 31, 1727.
(http://en.wikipedia.org/wiki/Isaac_Newton)
18
Analytical Questions:
Integrate:
5
1). ∫1 6𝑥 2 𝑑𝑥
10
2). ∫0 √𝑥 + 1𝑑𝑥
7 2
3). ∫5
𝑥2
𝑑𝑥
𝜋
4). ∫𝜋 cos 𝑥 𝑑𝑥
2
9
5). ∫3 sin 𝑥 + 𝑥 2 +
√𝑥
𝑑𝑥
2
Find 𝑓′(𝑥) using the Second Fundamental Theorem of Calculus
𝑥
6). 𝑓(𝑥) = ∫0 𝑡 2 𝑑𝑡
𝑥+2 2
7). 𝑓(𝑥) = ∫1
𝑡 + 5𝑡 + 2𝑑𝑡
3𝑥
8). 𝑓(𝑥) = ∫5 sin 𝑡 𝑑𝑡
3𝑥 2
9). 𝑓(𝑥) = ∫−2 4𝑡 3 𝑑𝑡
7𝑥
10). 𝑓(𝑥) = ∫0 𝑡𝑎𝑛2 𝑡 𝑑𝑡
AP Multiple Choice Questions
11). A is the area between the graphs of 𝑓(𝑥) = 𝑥 and 𝑔(𝑥) = 𝑥 3 from 0 to 1.
Find A.
a). .25
b). -.25
c). .20
19
d). 1
e). .5
12). Water is evaporating from a swimming pool at a rate 𝑅(𝑡) = 100𝑒 .01𝑡 gallons
per hour. If the pool holds 50000 gallons of water, when will it all evaporate?
a). 39.317 hours
b). 39.319 hours
c). 39.318 hours
d). 39.321 hours
e). 39.316 hours
4𝑥+1
13). If 𝐹(𝑥) = 3 ∫1
csc 𝑡 𝑑𝑡 + 𝑥, then 𝐹′(𝑥) =
a). 3 csc 𝑥 + 𝑥
b). 12 csc 4𝑥 + 4𝑥
c). 4(3 csc(4𝑥 + 1) + 4𝑥 + 1) + 1
d). 3 csc(4𝑥 + 1) + 4𝑥 + 1
e). 12 csc(4𝑥 + 1) + 4𝑥 + 1
14).
Hours:
h(t):
0
10
2
9
3
7
5
4
7
3
h(t) represents the change temperature of a cup of coffee sitting in a
freezer in Fahrenheit per hour. Using trapezoidal approximation, find the
temperature at t=7 if H(0) = 100 .
20
a). 55
b). 55.5
c). 60.1
d). 54.5
e). 55.1
3𝑥 3
13). If 𝐹(𝑥) = ∫0
2
5𝑡 3 𝑑𝑡, then 𝐹′(𝑥) =
a). (15𝑥 2 )
b). 15𝑥 5
c). 30𝑥 4
d). 405𝑥 4
e). 45𝑥 4
Practice FRQ
A factory in the middle of Yugoslavia manufactures cars through biodegradable material (I know, what?
How?). Anyway, they get biodegradable material from an outside agency. The rate at which the material
comes in pounds to the factory can be expressed by the equation 𝑔(𝑥)
18𝑥
= 3 cos (
7𝜋
) − 𝜋, where
x is equal to the number of days since the start of the calendar year. 50 days since the start of the
21
calendar year, there are 60 pounds of biodegradable material in the factory. 60 days after the start of
the calendar year, the administrators agreed to take material out of the factory. Starting on day 60 after
the start of the calendar year, the rate at which biodegradable material flows out of the factory can be
𝑒
expressed with the equation ℎ (𝑥 ) = sin( ). The factory continues to bring in biodegradable
2𝑥
throughout the calendar year.
a) With correct units, calculate how much biodegradable material is in the factory 30 days after the
start of the calendar year.
b) With correct units, calculate how much biodegradable material is in the factory 75 days after the
start of the calendar year
c) After the executive decision to remove biodegradable material each year, which was 60 days
after the start of the calendar year, on what day is there the most biodegradable material in the
factory, and what is that value? (There are 365 days in a year)
22