Antiderivatives

Definition A function F is called anantiderivative of f on an interval I if F ′(x) = f (x)for all x in I .

Example Let f (x) = x2. Then an antiderivativeF (x) for x2 is F (x) = x3

3 .

Theorem If F is an antiderivative of f on aninterval I , then the most general antiderivative of fon I is

F (x) + C

where C is an arbitrary constant.

Table of Antidifferentiation Formulas

Function Particular antiderivative

c · f (x) c · F (x)

f (x) + g(x) F (x) + G (x)

xn(n 6= −1) xn+1

n+11x ln |x |ex ex

cos x sin x

Table of Antidifferentiation Formulas

Function Particular antiderivative

sin x − cos x

sec2 x tan x

sec x tan x sec x1√1−x2

sin−1 x

11+x2 tan−1 x

Example A ball is thrown upward with a speed of48 ft/s from the edge of a cliff 432 ft above theground. Find its height above the ground t secondslater. When does it reach its maximum height?When does it hit the ground?

Solution The motion is vertical and we choose thepositive direction to be upward. At time t thedistance above the ground is s(t) and the velocityv(t) is decreasing.

Therefore, the acceleration must be negative and

a(t) =dv

dt= −32

Taking antiderivatives

v(t) = −32t + C

To determine C we use that v(0) = 48. This gives48 = 0 + C , so

v(t) = −32t + 48

The maximum height is reached when v(t) = 0,that is after 1.5 s. Since s ′(t) = v(t), weantidifferentiate again and obtains(t) = −16t2 + 48t + D

Using the fact that s(0) = 432, we have432 = 0 + D and so

s(t) = −16t2 + 48t + 432

The expression for s(t) is valid until the ball hits theground. This happens when s(t) = 0, that is when

−16t2 + 48t + 432 = 0

or, equivalently,

t2 − 3t − 27 = 0

t =3± 3

√13

2We reject solution with the minus sign since it givesa negative value for t. Therefore, the ball hits theground after 3(1 +

√13)/2 ≈ 6.9 s.

Area under a graph

Area under y = x2 from 0 to 1.

Example Use rectangles to estimate the areaunder the parabola y = x2 from 0 to 1.

Area estimate using right and points

R4 =1

4·(

1

4

)2

+1

4·(

1

2

)2

+1

4·(

3

4

)2

+1

4· 12 =

15

32

Note area A < 1532 = .46875

Area estimate using left end points

L4 =1

4·02+

1

4·(

1

4

)+

1

4·(

1

2

)2

+1

4·(

3

4

)2

=7

32= .21875

Note area satisfies

.21875 ≤ A ≤ .46875

General calculation using right end points

Area definition using right end points

Definition The area A of the region S that liesunder the graph of the continuous function f is thelimit of the sum of the areas of approximatingrectangles:

A = limn→∞

Rn

= limn→∞

[f (x1)∆x + f (x2)∆x + · · ·+ f (xn)∆x ]

Note in the above definition that if I = [a, b], then

∆x =b − a

n,

where n is the number of rectangles or divisions

Sigma notation and A = Area

n∑i=1

f (xi)∆x = f (x1)∆x + f (x2)∆x + · · ·+ f (xn)∆x

A = limn→∞

n∑i=1

f (xi)∆x

= limn→∞

n∑i=1

f (xi−1)∆x

= limn→∞

n∑i=1

f (x∗i )∆x , x∗i ∈ [xi−1, xi ].

Definition of a Definite Integral

Definition If f is a continuous function definedfor a ≤ x ≤ b, we divide the interval [a, b] into nsubintervals of equal width ∆x = (b − a)/n. We letx0(= a), x1, x2, . . . , xn(= b) be the endpoints ofthese subintervals and we let x∗1 , x

∗2 , . . . , x

∗n be any

sample points in these subintervals, so x∗i lies inthe i -th subinterval [xi−1,xi

]. Then the definiteintegral of f from a to b is∫ b

a

f (x) dx = limn→∞

∑i=1

f (x∗i )∆x

Midpoint Rule

∫ b

a

f (x) dx ≈n∑

i=1

f (xi)∆x = ∆x [f (x1)+· · ·+f (xn)]

where

∆x =b − a

nand

xi =1

2(xi−1 + xi) = midpoint of [xi−1, xi ]

Properties of the Integral

1.∫ b

a c dx = c(b − a), where c is any constant

2.∫ b

a [f (x) + g(x)] =∫ b

a f (x) dx +∫ b

a g(x) dx

3.∫ b

a c · f (x) dx = c∫ b

a f (x) dx , where c is anyconstant

4.∫ b

a [f (x)− g(x)] dx =∫ b

a f (x) dx −∫ b

a g(x) dx

The Fundamental Theorem of Calculus, Part 1

If f is continuous on [a, b], then the function

g defined by

g(x) =

∫ x

a

f (t) dt a ≤ x ≤ b

is continuous on [a, b] and differentiable on

(a, b), and

g ′(x) = f (x).

The Fundamental Theorem of Calculus, Part 2

If f is continuous on [a, b], then∫ b

a

f (x) dx = F (b)− F (a),

where F (x) is any antiderivative of f (x),

that is, a function such that

F ′(x) = f (x).

Application of FTC

Example Evaluate∫ 6

31x dx .

Solution An antiderivative for 1x is

F (x) = ln x . So, by FTC,∫ 6

3

1

xdx = F (6)− F (3) = ln 6− ln 3.

Example Evaluate∫ 3

1 ex dx .

Solution Note that an antiderivative for ex

is F (x) = ex . So, by FTC,∫ 3

1

ex dx = F (3)− F (1) = e3 − e.

Example Find area A under the cosine curvefrom 0 to b, where 0 ≤ b ≤ π

2 .

Solution Since an antiderivative of f (x) = cos(x) isF (x) = sin(x), we have

A =

∫ b

0

cos(x) dx = sin(x)

]b

0

= sin(b)−sin(0) = sin(b).

The Fundamental Theorem of Calculus

Suppose f is continuous on [a, b].

1. If g(x) =∫ x

a f (t) dt, theng ′(x) = f (x).

2.∫ b

a f (x) dx = F (b)− F (a), whereF (x) is any antiderivative of f (x),that is, F ′(x) = f (x).

Notation: Indefinite integral

∫f (x) dx = F (x) means F ′(x) = f (x).

We use the notation∫

f (x) dx to denote an

antiderivative for f (x) and it is called anindefinite integral. A definite integral has

the form:∫ b

a

f (x) dx =

∫f (x) dx

]b

a

= F (b)− F (a)

Table of Indefinite Integrals

∫c · f (x) dx = c ·

∫f (x) dx∫

[f (x) + g(x)] dx =∫

f (x) dx +∫

g(x) dx∫k dx = kx + C∫

xn dx = xn+1

n+1 + C (n 6= −1)∫1x dx = ln |x | + C∫ex dx = ex + C

Table of Indefinite Integrals

∫sin x dx = − cos x + C∫cos x dx = sin x + C∫sec2 x dx = tan x + C∫

csc2 x dx = − cot x + C∫sec x tan x dx = sec x + C∫

csc x cot x dx = − csc x + C∫1

x2+1dx = tan−1 x + C∫

1√1−x2

dx = sin−1 x + C

Example Find∫ 2

0

(2x3 − 6x + 3

x2+1

)dx and

interpret the result in terms of areas.Solution FTC gives∫ 2

0

(2x3 − 6x +

3

x2 + 1

)dx

= 2x4

4− 6

x2

2+ 3 tan−1 x

]2

0

.

=1

2x4 − 3x2 + 3 tan−1 x

]2

0

=1

2(24)− 3(22) + 3 tan−1 2− 0

= −4 + 3 tan−1 2

The Net Change Theorem

The Net Change Theorem The integral

of a rate of change is the net change:∫ b

a

F ′(x) dx = F (b)− F (a).

Example A particle moves along a line so that itsvelocity at time t is v(t) = t2 − t − 6 (measured inmeters per second). Find the displacement of theparticle during the time period 1 ≤ t ≤ 4.Solution By the Net Change Theorem, thedisplacement is

s(4)− s(1) =

∫ 4

1

v(t)dt =

∫ 4

1

(t2 − t − 6) dt

=

[t3

3− t2

2− 6t

]4

1

= −9

2.

This means that the particle moved 4.5 m towardthe left.

Example A particle moves along a line so that itsvelocity at time t is v(t) = t2 − t − 6 (measured inmeters per second). Find the distance traveledduring the time period 1 ≤ t ≤ 4.Solution Note that v(t) = t2 − t − 6 = (t − 3)(t + 2) andso v(t) ≤ 0 on the interval [1, 3] and v(t) ≥ 0 on [3, 4]. Fromthe Net Change Theorem, the distance traveled is∫ 4

1

|v(t)| dt =

∫ 3

1

[−v(t)] dt +

∫ 4

3

v(t) dt

=

∫ 3

1

(−t2 + t + 6) dt +

∫ 4

3

(t2 − t − 6) dt

=

[−t3

3+

t2

2+ 6t

]3

1

+

[t3

3− t2

2− 6t

]4

3

=61

6≈ 10.17m

The Substitution Rule

The Substitution Rule is one of the main

tools used in this class for finding

antiderivatives. It comes from the Chain

Rule:

[F (g(x))]′ = F ′(g(x))g ′(x).

So, ∫F ′(g(x))g ′(x) dx = F (g(x)).

The Substitution Rule

The Substitution Rule If u = g(x) is a differentiablefunction whose range is an interval I and f is continuous on I ,then ∫

f (g(x))g ′(x) dx =

∫f (u) du.

Example Find∫

x3 cos(x4 + 2) dx .Solution

1. Make the substitution: u = x4 + 2.

2. Get du = 4x3 dx .∫x3 cos(x4 + 2) dx =

∫cos u · 1

4du =

1

4

∫cos u du

=1

4sin u + C =

1

4sin(x4 + 2) + C .

Note at the final stage we return to the original variable x .

The Substitution Rule If u = g(x) is a differentiablefunction whose range is an interval I and f is continuous on I ,then ∫

f (g(x))g ′(x) dx =

∫f (u) du.

Example Evaluate∫ √

2x + 1 dx .Solution Let u = 2x + 1. Then du = 2 dx , so dx = du

2. Thus,

the Substitution Rule gives∫ √2x + 1 dx =

∫ √u

du

2=

1

2

∫u

12 du

=1

2· u

32

32

+ C =1

3u

32 + C =

1

3(2x + 1)

32 + C .

The Substitution Rule If u = g(x) is a differentiablefunction whose range is an interval I and f is continuous on I ,then ∫

f (g(x))g ′(x) dx =

∫f (u) du.

Example Calculate∫

e5x dx .Solution If we let u = 5x , then du = 5 dx , so dx = 1

5du

Therefore∫e5x dx =

1

5

∫eu du =

1

5eu + C =

1

5e5x + C .

The Substitution Rule If u = g(x) is a differentiablefunction whose range is an interval I and f is continuous on I ,then ∫

f (g(x))g ′(x) dx =

∫f (u) du.

Example Calculate∫

tan x dx .Solution ∫

tan x dx =

∫sin x

cos xdx .

This suggests substitution u = cos x , since thendu = − sin x dx and so, sin x dx = −du:∫

tan x dx =

∫sin x

cos xdx = −

∫du

u

= − ln |u|+ C = − ln | cos x |+ C .

The Substitution Rule for Definite Integrals

Substitution Rule for Definite IntegralsIf g ′ is continuous on [a, b] and f is

continuous on the range of u = g(x), then∫ b

a

f (g(x))g ′(x) dx =

∫ g(b)

g(a)

f (u) du.

The Substitution Rule for Definite IntegralsIf g ′ is continuous on [a, b] and f is continuous on the rangeof u = g(x), then∫ b

a

f (g(x))g ′(x) dx =

∫ g(b)

g(a)

f (u) du.

Example Calculate∫ e

1ln xx

dx .

Solution We let u = ln x because its differential du = dxx

occurs in the integral. When x = 1, u = ln 1 = 0; when x = e,u = ln e = 1. Thus∫ e

1

ln x

xdx =

∫ 1

0

u du =u2

2

]1

0

=1

2.