Fundamental Theorem of Calculus
P. Sam Johnson
April 10, 2017
P. Sam Johnson
Fundamental Theorem of Calculus
April 10, 2017
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Overview
In this lecture we present the Fundamental Theorem of Calculus, which
is the central theorem of integral calculus.
It connects integration and differentiation, enabling us to compute
integrals using an antiderivative of the integrand function rather than by
taking limits of Riemann sums.
Leibniz and Newton exploited this relationship and started mathematical
developments that fueled the scientific revolution for the next 200 years.
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Recall
Theorem (The Mean Value Theorem)
If f is continuous on a closed interval [a, b] and f is differentiable on the
interval’s interior (a, b). Then there is at least one point c in (a, b) at
which
f (b) − f (a)
= f 0 (c).
b−a
Geometrically, there is a point where the tangent is parallel to the chord joining (a, f (a)) and
(b, f (b)).
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Recall
Theorem (The Intermediate Value Theorem for Continuous
Functions)
A function y = f (x) that is continuous on a closed interval [a, b] takes on
every value between f (a) and f (b).
In other words, if y0 is any value between f (a) and f (b), then
y0 = f (c)
for some c in [a, b].
Geometrically, the Intermediate Value Theorem
says that any horizontal line y = y0 crossing the
y -axis between the numbers f (a) and f (b) will
cross the curve y = f (x) at least once over the
interval [a, b].
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The Mean Value Theorem for Definite Integrals
First we present the integral version of the Mean Value Theorem,
which is another important theorem of integral calculus and used to
prove the Fundamental Theorem.
The Mean Value Theorem for Definite Integrals asserts that the average
value is always taken on at least once by the function f in the interval.
Geometrically, the Mean Value Theorem says that
there is a number c in [a, b] such that the rectangle with height equal to the average value f (c)
of the function and base width b − a has exactly
the same area as the region beneath the graph of
f from a to b.
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Theorem (The Mean Value Theorem for Definite Integrals)
If f is continuous on [a, b], then at some point c in [a, b],
Z b
1
f (c) =
f (x) dx.
b−a a
Proof of the theorem : If we divide both sides of the Max-Min
Inequality by (b − a), we obtain
Z b
1
min f ≤
f (x) dx ≤ max f .
b−a a
Since f is continuous, the Intermediate Value Theorem for Continuous
Functions says that f must assume every value between min f and max f .
It must therefore assume the value
Z b
1
f (x) dx
b−a a
at some point c in [a, b].
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Exercises
1. Show that if f is continuous on [a, b], a 6= b, and if
Z b
f (x) dx = 0,
a
then f (x) = 0 at least once in [a, b].
2. Determine the number c that satisfies the Mean Value Theorem for
Definite integrals for the function
f (x) = x 2 + 3x + 2
on [1, 4].
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Theorem (Fundamental Theorem of Calculus - Part 1)
If f is continuous on [a, b] then
Z
F (x) =
x
f (t) dt
a
is continuous on [a, b] and differentiable on (a, b) and its derivative is
f (x). That is,
Z x
d
0
F (x) =
f (t) dt = f (x).
dx a
Proof of the theorem : If f (t) is an integrable function over a finite
integral I , then the integral from any fixed number a ∈ I to another
number x ∈ I defines a new function F whose value at x is
Z x
f (t) dt.
F (x) =
a
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For x, x + h ∈ (a, b), we have
Z x+h
Z
F (x + h) − F (x) =
f (t) dt −
a
x
Z
x+h
f (t) dt =
a
f (t) dt
x
by additivity rule for integrals.
Hence
F (x + h) − F (x)
1
=
h
h
Z
x+h
f (t) dt.
(1)
x
By the Mean Value Theorem for Definite Integrals, there is some c in this
interval such that
Z
1 x+h
f (t) dt = f (c).
h x
As h → 0, x + h approaches x, forcing c to approach x also.
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Since f is continuous at x, f (c) approaches f (x) as h → 0. Hence
dF
dx
F (x + h) − F (x)
h→0
h
Z x+h
1
f (t) dt
= lim
h→0 h x
= lim f (c) = f (x).
=
lim
h→0
If x = a or b, then the limit of Equation (1) is interpreted as a one-sided
limit with h → 0+ or h → 0− , respectively.
This shows that F is continuous for every point of [a, b].
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Fundamental Theorem of Calculus - Part 2
(The Evaluation Theorem)
Theorem
If f is continuous at every point of [a, b] and F is any antiderivative of f
on [a, b], then
Z b
f (x) dx = F (b) − F (a).
a
Proof of the theorem : Part 1 of the Fundamental Theorem tells us that
an antiderivative of f exists, namely
Z x
G (x) =
f (t) dt.
a
Thus, if F is any antiderivative of f , then
F (x) = G (x) + C
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Johnson
Fundamental
for
some
constant C for a <
x < b.Theorem of Calculus
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Since both F and G are continuous on [a, b], we see that
F (x) = G (x) + C
also holds when x = a and x = b by taking one-sided limits (as x → a+
and x → b− ).
Evaluating F (b) − F (a), we have
F (b) − F (a) = [G (b) + C ] − [G (a) + C ]
= G (b) − G (a)
Z b
Z a
=
f (t) dt −
f (t) dt
a
a
Z b
=
f (t) dt − 0
a
Z b
=
f (t) dt.
a
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Evaluation Theorem
The Evaluation Theorem is important because it says that to calculate the
definite integral of f over an interval [a, b] we need do only two things:
(a) Find an antiderivative F of f , and
(b) Calculate the number F (b) − F (a), which is equal to
Rb
a
f (x) dx.
This process is much easier then using a Riemann sum computation.
The power of the theorem follows from the realization that the definite
integral, which is defined by a complicated process involving all of the
values of the function f over [a, b], can be found by knowing the values of
any antiderivative F at only the two endpoints a and b.
The usual notation for the difference F (b) − F (a) is
ib
h
ib
F (x)
or F (x)
a
a
depending on whether F has one or more terms.
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Evaluation Theorem
Example
We calculate several definite integrals using the Evaluation Theorem,
rather than by taking limits of Riemann sums.
Z π
iπ
d
cos xdx = sin x
sin x = cos x
dx
0
0
= sin π − sin 0
= 0−0
= 0.
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The Integral of a Rate
We can interpret Part 2 of the Fundamental Theorem in another way. If F
is any antiderivative of f , then F 0 = f . The equation in the theorem can
then be rewritten as
Z b
F 0 (x) dx = F (b) − F (a).
a
F 0 (x)
Now
represents the rate of change of the function F (x) with respect
to x, so the integral of F 0 is just the net change in F as x changes from a
to b. Formally, we have the following result.
Theorem (The Net Change Theorem)
The net change in a function F (x) over an interval a ≤ x ≤ b is the
integral of its rate of change :
Z b
F (b) − F (a) =
F 0 (x) dx.
a
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Displacement over the time interval
If an object with position function s(t) moves along a coordinate line, its
velocity is v (t) = s 0 (t). Net Change Theorem says that
Z t2
v (t) dt = s(t2 ) − s(t1 ),
t1
so the integral of velocity is the displacement over the time interval
t1 ≤ t ≤ t2 .
On the other hand, the integral of the speed
|v (t)|
is the total distance traveled over the time interval.
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Displacement over the time interval
As
Z
F (b) = F (a) +
b
F 0 (x) dx
a
we see that the Net Change Theorem also says that the final value of a
function F (x) over an interval [a, b] equals its initial value F (a) plus its
net change over the interval.
So if v (t) represents the velocity function of an object moving along a
coordinate line, this means the the object’s final position s(t2 ) over a time
interval
t1 ≤ t ≤ t2
is its initial position s(t1 ) plus its net change in position along the line.
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The Relationship between Integration and Differentiation
The conclusions of theZ Fundamental Theorem tell us several things. The
x
d
equation F 0 (x) =
f (t) dt = f (x) can be rewritten as
dx a
Z x
d
dF
f (t) dt =
= f (x),
dx a
dx
which says that if we first integrate the function f and then
differentiate the result, we get the function f back again.
Likewise, the equation
Z x
Z x
dF
f (t) dt =
f (t) dt = F (x) − F (a)
a dt
a
says that if we first differentiate the function F and then integrate
the result, we get the function F back (adjusted by an integration
constant).
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The Relationship between Integration and Differentiation
In a sense, the processes of integration and differentiation are
“inverses” of each other.
The Fundamental Theorem also says that every continuous function f
has an antiderivative F .
And it says that the differential equation
dy
= f (x)
dx
has a solution (namely, the function y = F (x)) for every continuous
function f .
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Total Area
To compute the area of the region by the graph of a function y = f (x)
and the x-axis requires more care when the function takes on both positive
and negative values.
We must be careful to break up the interval [a, b] into subintervals on
which the function doesn’t change sign.
Otherwise we might get cancellation between positive and negative signed
areas, leading to an incorrect total.
The correct total area is obtained by adding the absolute value of the
definite integral over each subinterval where f (x) does not change sign.
The term “area” will be taken to mean total area.
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Summary
To find the area between the graph of y = f (x) and the x-axis over the
interval [a, b], do the following:
1. Subdivide [a, b] at the zeros of f .
2. Integrate f over each subinterval.
3. Add the absolute values of the integrals.
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Exercises
3. Evaluate the following integrals.
Z
5π/6
csc2 x dx
(a)
π/6
Z −π/4
(b)
−π/3
√
3
Z
(c)
√
Z−π
(d)
0
P. Sam Johnson
4 sec2 t +
π
dt
t2
(t + 1)(t 2 + 4) dt
3
1
(cos x + | cos x|) dx.
2
Fundamental Theorem of Calculus
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Exercises
Find the derivatives of the following.
(a) by evaluating the integral and differentiating the result.
(b) by differentiating the integral directly.
Z √x
d
4.
cos t dt
dx 0
Z tan θ
d
5.
sec y dy
dθ 0
Z t4
√
d
6.
u du.
dt 0
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Exercises
7. Find
dy
dx
of the following.
Z
x
(a) y =
Z0 x
(b) y =
Z1 0
(c) y =
√
Z
(d) y =
1
p
1 + t 2 dt
1
dt
t
sin(t 2 ) dt
x
sin x
√
dt
,
1 − t2
|x| < π/2
8. Find the total area between the region and the x-axis
(a) y = −x 2 − 2x, −3 ≤ x ≤ 2
(b) y = x 3 − 4x, −2 ≤ x ≤ 2
(c) y = x 1/3 − x, −1 ≤ x ≤ 8.
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Exercises
9. Find the areas of the shaded regions of the following.
(a)
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(b)
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Exercises
10. Each of the following functions solves one of the initial value
problems. Which function solves which problem? Give brief reasons
for your answers.
Initial value problems
Z
x
(a) y =
Z1 x
(b) y =
1
dt
t
sec t dt + 4
Z0
1
(c) y =
sec t dt
Z−1
x
(d) y =
π
1
dt − 3
t
Solutions
(a) dy
y (2) = 3
dx = sec x,
(b) y 0 = sec x, y (−1) = 4
(c) y 0 = sec x, y (0) = 4
(d) y 0 = x1 , y (1) = −3.
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Exercises
11. Express the solutions of the initial value problem in terms of integrals.
dy
dx
dy
(b)
dx
ds
(c)
dt
dv
(d)
dt
(a)
P. Sam Johnson
= sec x, y (2) = 3
p
= 1 + x 2 , y (1) = −2
= f (t),
s(t0 ) = s0
= g (t),
v (t0 ) = v0 .
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Archimedes’ area formula for parabolas
Exercises
12. Archimedes (287-212 B.C.), inventor, military engineer, physicist, and
the greatest mathematician of classical times in the Western world,
discovered that the area under a parabolic arch is two-thirds the base
times the height. Sketch the parabolic arch
y = h − (4h/b 2 )x 2
− b/2 ≤ x ≤ b/2
assuming that h and b are positive. Then use calculus to find the
area of the region enclosed between the arch and the x-axis.
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Exercises
13. Consider a heavy rock blown straight up from the ground by a
dynamite blast. The velocity of the rock at any time t during the
motion was given as v (t) = 160 − 32t ft/sec.
(a) Find the displacement of the rock during the time period 0 ≤ t ≤ 8.
(b) Find the total distance traveled during this time period.
14. Revenue from marginal revenue: Suppose that a company’s
marginal revenue from the manufacture and sale of eggbeaters is
2
dr
=2−
,
dt
(x + 1)2
where r is measured in thousands of dollars and x in thousands of
units. How much money should the company expect from a
production run of x = 3 thousand eggbeaters? To find out, integrate
the marginal revenue from x = 0 to x = 3.
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Exercises
15. The temperature T (◦ F ) of a room at time t minutes is given by
√
T = 85 − 3 25 − t for 0 ≤ t ≤ 25.
(a) Find the room’s temperature when t = 0, t = 16, and t = 25.
(b) Find the room’s average temperature for 0 ≤ t ≤ 25.
16. The height H (ft) of a palm tree after growing for t years is given by
√
H = t + 1 + 5t 1/3 for 0 ≤ t ≤ 8.
(a) Find the tree’s height when t = 0, t = 4, and t = 8.
(b) Find the tree’s average height for 0 ≤ t ≤ 8.
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Exercises
Z
x
17. Suppose that
f (t) dt = x 2 − 2x + 1. Find f (x).
1
Z x
18. Find f (4) if
f (t) dt = x cos πx.
0
Z x
Z 1
19. If
f (t) dt = x +
f (t) dt, then find the value of f (1).
0
0
Z x
20. Suppose
f (t) dt = x 2 (1 + x), x ≥ 0 find the value of f (2).
0
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Table of Integrals : Basic Forms
Z
x n dx =
Z
1
dx = ln |x|
x
1
x n+1 , n 6= −1
n+1
Z
Z
udv = uv −
Z
vdu
1
1
dx = ln |ax + b|
ax + b
a
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Integrals of Rational Functions
Z
Z
Z
Z
Z
Z
1
1
dx = −
(x + a)2
x +a
(x + a)n+1
(x + a)n dx =
, n 6= −1
n+1
(x + a)n+1 ((n + 1)x − a)
x(x + a)n dx =
(n + 1)(n + 2)
1
1
x
dx = tan−1
2
2
a +x
a
a
x
1
dx = ln |a2 + x 2 |
a2 + x 2
2
2
x
x
dx = x − a tan−1
a2 + x 2
a
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Integrals of Rational Functions
x3
1
1
dx = x 2 − a2 ln |a2 + x 2 |
2
2
a +x
2
2
Z
1
2
2ax + b
dx = √
tan−1 √
2
2
ax + bx + c
4ac − b
4ac − b 2
Z
1
1
a+x
dx =
ln
, a 6= b
(x + a)(x + b)
b−a b+x
Z
1
dx = tan−1 x
1 + x2
Z
x
a
dx =
+ ln |a + x|
(x + a)2
a+x
Z
x
dx =
2
ax + bx + c
1
b
2ax + b
ln |ax 2 + bx + c| − √
tan−1 √
2a
a 4ac − b 2
4ac − b 2
Z
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Integrals with Roots
Z
Z
Z
Z
Z
Z
Z
√
2
x − a dx = (x − a)3/2
3
√
1
√
dx = 2 x ± a
x ±a
√
2b 2x √
+
ax + b dx =
ax + b
3a
3
2
(ax + b)3/2 dx = (ax + b)5/2
5a
p
r
p
x(a − x)
x
−1
dx = − x(a − x) − a tan
a−x
x −a
r
p
√
√
x
dx = x(a + x) − a ln x + x + a
a+x
√
√
2
x ax + b dx =
(−2b 2 + abx + 3a2 x 2 ) ax + b
2
15a
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Integrals with Roots
Z p
1 p
1
x
a2 − x 2 dx = x a2 − x 2 + a2 tan−1 √
2
2
2
a − x2
Z p
1 2
3/2
x x 2 ± a2 dx =
x ± a2
3
Z
p
1
√
dx = ln x + x 2 ± a2 2
2
x ±a
Z
1
x
√
dx = sin−1
2
2
a
a −x
Z
p
x
√
dx = x 2 ± a2
2 ± a2
x
Z
p
x
√
dx = − a2 − x 2
a2 − x 2
Z
p
x2
1 p
1
√
dx = x x 2 ± a2 ∓ a2 ln x + x 2 ± a2 2
2
x 2 ± a2
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Integrals with Logarithms
Z
ln ax dx = x ln ax − x
x2
1
x ln x dx = x 2 ln x −
2
4
Z
1
x3
x 2 ln x dx = x 3 ln x −
3
9
Z
1
ln x
n
n+1
x ln x dx = x
−
,
n + 1 (n + 1)2
Z
ln ax
1
dx = (ln ax)2
x
2
Z
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Integrals with Logarithms
Z
ln x
1 ln x
dx = − −
x2
x
x
Z
b
ln(ax + b) dx = x +
ln(ax + b) − x, a 6= 0
a
Z
x
ln(x 2 + a2 ) dx = x ln(x 2 + a2 ) + 2a tan−1 − 2x
a
Z
x
+
a
ln(x 2 − a2 ) dx = x ln(x 2 − a2 ) + a ln
− 2x
x −a
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Integrals with Exponentials
Z
1
e ax dx = e ax
a
Z
xe x dx = (x − 1)e x
1
x
ax
−
xe dx =
e ax
a a2
x 2 e x dx = x 2 − 2x + 2 e x
2
x
2x
2
2 ax
x e dx =
− 2 + 3 e ax
a
a
a
x 3 e x dx = x 3 − 3x 2 + 6x − 6 e x
Z
x n e ax
n
n ax
x e dx =
−
x n−1 e ax dx
a
a
Z
Z
Z
Z
Z
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Integrals with Trigonometric Functions
Z
Z
Z
Z
Z
Z
Z
Z
1
sin ax dx = − cos ax
a
sin 2ax
x
sin2 ax dx = −
2
4a
3
cos
ax
cos 3ax
sin3 ax dx = −
+
4a
12a
1
cos ax dx = sin ax
a
x
sin 2ax
cos2 ax dx = +
2
4a
sin 3ax
3
sin
ax
+
cos3 axdx =
4a
12a
1 2
1
1
cos x sin x dx = sin x + c1 = − cos2 x + c2 = − cos 2x + c3
2
2
4
cos[(a − b)x] cos[(a + b)x]
cos ax sin bx dx =
−
, a 6= b
2(a − b)
2(a + b)
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Integrals with Trigonometric Functions
Z
sin2 ax cos bx dx = −
Z
sin2 x cos x dx =
Z
Z
Z
Z
Z
sin[(2a − b)x] sin bx
sin[(2a + b)x]
+
−
4(2a − b)
2b
4(2a + b)
1 3
sin x
3
cos[(2a − b)x] cos bx
cos[(2a + b)x]
cos2 ax sin bx dx =
−
−
4(2a − b)
2b
4(2a + b)
1
cos2 ax sin ax dx = − cos3 ax
3a
x
sin 4ax
sin2 ax cos2 ax dx = −
8
32a
1
tan ax dx = − ln cos ax
a
1
tan2 ax dx = −x + tan ax
a
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Integrals with Trigonometric Functions
Z
tan3 axdx =
1
1
ln cos ax +
sec2 ax
a
2a
Z
sec x dx = ln | sec x + tan x| = 2 tanh
Z
−1
x
tan
2
1
tan ax
a
Z
1
1
sec3 x dx = sec x tan x + ln | sec x + tan x|
2
2
Z
sec x tan x dx = sec x
Z
1
sec2 x tan x dx = sec2 x
2
Z
1
secn x tan x dx = secn x, n 6= 0
n
sec2 ax dx =
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Integrals with Trigonometric Functions
Z
x csc x dx = ln tan = ln | csc x − cot x| + C
2
Z
1
csc2 ax dx = − cot ax
a
Z
1
1
csc3 x dx = − cot x csc x + ln | csc x − cot x|
2
2
Z
1
cscn x cot x dx = − cscn x, n 6= 0
n
Z
sec x csc x dx = ln | tan x|
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Fundamental Theorem of Calculus
April 10, 2017
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Products of Trigonometric Functions and Monomials
Z
x cos x dx = cos x + x sin x
Z
1
x
cos ax + sin ax
a2
a
Z
x 2 cos x dx = 2x cos x + x 2 − 2 sin x
Z
2x cos ax
a2 x 2 − 2
x 2 cos ax dx =
+
sin ax
a2
a3
Z
x sin x dx = −x cos x + sin x
Z
x cos ax
sin ax
x sin ax dx = −
+
a
a2
x cos ax dx =
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Fundamental Theorem of Calculus
April 10, 2017
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Products of Trigonometric Functions and Monomials
Z
x 2 sin x dx = 2 − x 2 cos x + 2x sin x
2x sin ax
2 − a2 x 2
cos ax +
3
a
a2
Z
x2 1
1
x cos2 x dx =
+ cos 2x + x sin 2x
4
8
4
Z
2
x
1
1
x sin2 x dx =
− cos 2x − x sin 2x
4
8
4
Z
2
x
x tan2 x dx = − + ln cos x + x tan x
2
Z
x sec2 x dx = ln cos x + x tan x
Z
x 2 sin ax dx =
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Fundamental Theorem of Calculus
April 10, 2017
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Products of Trigonometric Functions and Exponentials
Z
Z
Z
Z
Z
Z
1
e x sin x dx = e x (sin x − cos x)
2
1
e bx sin ax dx = 2
e bx (b sin ax − a cos ax)
a + b2
1
e x cos x dx = e x (sin x + cos x)
2
1
e bx cos ax dx = 2
e bx (a sin ax + b cos ax)
a + b2
1
xe x sin x dx = e x (cos x − x cos x + x sin x)
2
1
xe x cos x dx = e x (x cos x − sin x + x sin x)
2
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Fundamental Theorem of Calculus
April 10, 2017
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Integrals of Hyperbolic Functions
Z
1
sinh ax
a
1
sinh ax dx = cosh ax
a
1
tanh ax dx = ln cosh ax
a
1
cos ax cosh bx dx = 2
[a sin ax cosh bx + b cos ax sinh bx]
a + b2
1
cos ax sinh bx dx = 2
[b cos ax cosh bx + a sin ax sinh bx]
a + b2
1
sin ax cosh bx dx = 2
[−a cos ax cosh bx + b sin ax sinh bx]
a + b2
1
sin ax sinh bx dx = 2
[b cosh bx sin ax − a cos ax sinh bx]
a + b2
1
sinh ax cosh axdx =
[−2ax + sinh 2ax]
4a
1
sinh ax cosh bx dx = 2
[b cosh bx sinh ax − a cosh ax sinh bx]
b − a2
cosh ax dx =
Z
Z
Z
Z
Z
Z
Z
Z
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Fundamental Theorem of Calculus
April 10, 2017
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References
1. M.D. Weir, J. Hass and F.R. Giordano, Thomas’ Calculus, 11th
Edition, Pearson Publishers.
2. R. Courant and F.John, Introduction to calculus and analysis, Volume
II, Springer-Verlag
3. N. Piskunov, Differential and Integral Calculus, Vol I & II (Translated
by George Yankovsky).
4. E. Kreyszig, Advanced Engineering Mathematics, Wiley Publishers.
P. Sam Johnson
Fundamental Theorem of Calculus
April 10, 2017
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