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

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 = du2

. 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 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 = 15du.

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∫

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 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.

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.

Areas between curves

Area between curves

The area A of the region bounded by the curvesy = f (x), y = g(x), and the lines x = a, x = b,where f and g continuous and f (x) ≥ g(x) for all xin [a, b], is

A =

∫ b

a

[f (x)− g(x)] dx

Example Find the area of the region boundedabove by y = ex , bounded below by y = x , andbounded on the sides by x = 0 and x = 1.

Example Find the area of the region bounded above by

y = ex , bounded below by y = x , and bounded on the sides

by x = 0 and x = 1.

Solution The region is shown in the figure. The upperboundary curve is y = ex and the lower boundary curve isy = x . So we use the area formula with f (x) = ex , g(x) = x ,a = 0, and b = 1:

A =

∫ 1

0

(ex − x) dx = ex − 1

2x2

]1

0

= e − 1

2− 1 = e − 1.5

Example Find the area of the region enclosed bythe parabolas y = x2 and y = 2x − x2.

Example Find the area of the region enclosed bythe parabolas y = x2 and y = 2x − x2.

Solution We first find the points of intersection of theparabolas by solving their equations simultaneously. This givesx2 = 2x − x2, or 2x2 − 2x = 0. Thus 2x(x − 1) = 0, so x = 0or 1. The points of intersection are (0, 0) and (1, 1). So thetotal area is:

A =∫ 1

0(2x − 2x2) dx = 2

∫ 1

0(x − x2) dx = 2

[x2

2− x3

3

]1

0

= 2(

12− 1

3

)= 1

3

Areas between curves

The area between the curves y = f (x) andy = g(x) and between x = a and x = b is

A =

∫ b

a

|f (x)− g(x)| dx

.

Example Find the area enclosed by the liney = x − 1 and the parabola y 2 = 2x + 6.

Example Find the area enclosed by the line y = x − 1

and the parabola y 2 = 2x + 6.

Volumes

Volumes

V = volume ≈∑n

i=1 A(x∗i )∆x

Definition of volume

Definition of Volume Let S be a solid that liesbetween x = a and x = b. If the cross-sectionalarea of S in the plane Px , through x andperpendicular to the x−axis, is A(x), where A is acontinuous function, then the volume of S is

V = limn→∞

n∑i=1

A(x∗i )∆x =

∫ b

a

A(x) dx

Computing volume of a sphere

Example Show the volume of a sphere of radius ris V = 4

3πr 3.Solution x2 + y 2 = r 2 y =

√r 2 − y 2.

So the crossectional area is A(x) = πy 2 = π(r 2 − x2). So,V =

∫ r

−rA(x) dx =

∫ r

−rπ(r 2 − x2) dx = 2π

∫ r

0(r 2 − x2) dx

= 2π[r 2x − x3

3

]r

0= 2π

(r 3 − r3

3

)= 4

3πr 3

Computing volume of a sphere

Example Show the volume of a sphere of radius ris V = 4

3πr 3.

Solution x2 + y 2 = r 2 y =√

r 2 − y 2.So the crossectional area is A(x) = πy 2 = π(r 2 − x2). So,V =

∫ r

−rA(x) dx =

∫ r

−rπ(r 2 − x2) dx = 2π

∫ r

0(r 2 − x2) dx

= 2π[r 2x − x3

3

]r

0= 2π

(r 3 − r3

3

)= 4

3πr 3

Computing volume of a sphere

Example Show the volume of a sphere of radius ris V = 4

3πr 3.Solution x2 + y 2 = r 2 y =

√r 2 − y 2.

So the crossectional area is A(x) = πy 2 = π(r 2 − x2). So,V =

∫ r

−rA(x) dx =

∫ r

−rπ(r 2 − x2) dx = 2π

∫ r

0(r 2 − x2) dx

= 2π[r 2x − x3

3

]r

0= 2π

(r 3 − r3

3

)= 4

3πr 3

Computing a volume of revolution

Example Find the volume of the solid obtained by

rotating about the x-axis the region under the curve y =√

x

from 0 to 1.

Solution The area of the crossection is:A(x) = π(

√x)2 = πx .

So, V =∫ 1

0A(x) dx =

∫ 1

0πx dx = π x2

2

]1

0= π

2

Computing a volume of revolution

Example Find the volume of the solid obtained by

rotating about the x-axis the region under the curve y =√

x

from 0 to 1.

Solution The area of the crossection is:A(x) = π(

√x)2 = πx .

So, V =∫ 1

0A(x) dx =

∫ 1

0πx dx = π x2

2

]1

0= π

2

Volume of a solid paraboloid

Example Find the volume of the solid obtained by

rotating the region bounded by y = x3, y = 8 and x = 0

about the y -axis.

Volume of a solid paraboloid

Example Find the volume of the solid obtained by

rotating the region bounded by y = x3, y = 8 and x = 0

about the y -axis.

Solution Note that x = 3√

y . The area of crossection is:

A(y) = πx2 = π( 3√

y)2 = πy23 . So,

V =∫∞

0A(y) dy =

∫∞0

πy23

dy = π[

35y

53

]8

0= 96π

5.

Other volumes

Example The region R enclosed by the curves y = x and

y = x2 is rotated about the x-axis. Find the volume of the

solid region.

Solution The curves y = x and y = x2 intersect at thepoints (0, 0) and (1, 1). Crossection of rotated surface has theshape of a washer (annular ring). So the crossectional area is:A(x) = πx2 − π(x2)2 = π(x2 − x4).

So, V =∫ 1

0A(x) dx =

∫ 1

0π(x2 − x2) dx = π

[x3

3− x5

5

]= 2π

15.

Other volumes

Example The region R enclosed by the curves y = x and

y = x2 is rotated about the x-axis. Find the volume of the

solid region.

Solution The curves y = x and y = x2 intersect at thepoints (0, 0) and (1, 1). Crossection of rotated surface has theshape of a washer (annular ring). So the crossectional area is:A(x) = πx2 − π(x2)2 = π(x2 − x4).

So, V =∫ 1

0A(x) dx =

∫ 1

0π(x2 − x2) dx = π

[x3

3− x5

5

]= 2π

15.

Example The region R enclosed by the curves y = x and

y = x2 is rotated about the line y = 2. Find the volume.

Solution The crossection is again a washer. Thecrossectional area is:A(x) = π(2− x2)2 − π(2− x)2 = π(x4 − 5x2 + 4x).

So, V =∫ 1

0A(x) dx = π

∫ 1

0(x4 − 5x2 + 4x) dx =

π[

x5

5− 5 x3

3+ 4 x2

2

]1

0= 8π

15.

Example The region R enclosed by the curves y = x and

y = x2 is rotated about the line y = 2. Find the volume.

Solution The crossection is again a washer. Thecrossectional area is:A(x) = π(2− x2)2 − π(2− x)2 = π(x4 − 5x2 + 4x).

So, V =∫ 1

0A(x) dx = π

∫ 1

0(x4 − 5x2 + 4x) dx =

π[

x5

5− 5 x3

3+ 4 x2

2

]1

0= 8π

15.

Integration by parts

We now return to integration methods. Recall ourfirst method was substraction which came from thechain rule. Integration by parts comes from theproduct rule.

d

dx[f (x)g(x)] = f (x)g ′(x) + g(x)f ′(x).

So,

f (x)g ′(x) = (f (x)g(x))′ − g(x)f ′(x)

Thus, after taking antiderivatives, we get∫f (x)g ′(x) = f (x)g(x)−

∫g(x)f ′(x) dx .

Integration by parts

∫f (x)g ′(x) dx = f (x)g(x)−

∫g(x)f ′(x) dx

The above is called the formula for integrationby parts.

If we let u = f (x) and v = g(x), then du = f ′(x) dxand dv = g ′(x) dx , so the formula becomes:∫

u dv = uv −∫

v du.

Integration by parts

∫f (x)g ′(x) dx = f (x)g(x)−

∫g(x)f ′(x) dx

The above is called the formula for integrationby parts.If we let u = f (x) and v = g(x), then du = f ′(x) dxand dv = g ′(x) dx , so the formula becomes:∫

u dv = uv −∫

v du.

Strategy for using integration by parts

Recall the integration by parts formula:∫u dv = uv −

∫v du.

To apply this formula we must choose dv so that wecan integrate it! Frequently, we choose u so that thederivative of u is simpler than u. If both propertieshold, then you have made the correct choice.

Examples using strategy

1.

∫x sin x dx : Choose u = x and dv = sin x dx

2.

∫xex dx : Choose u = x and dv = ex dx

3.

∫t2et dt : Choose u = t2 and dv = et dt

4.∫

x2 sin 2x dx : Choose u = x2 and v = sin 2x dx

5.

∫ln(x) dx : Choose u = ln x and v = dx

Examples using strategy

1.

∫x sin x dx : Choose u = x and dv = sin x dx

2.

∫xex dx : Choose u = x and dv = ex dx

3.

∫t2et dt : Choose u = t2 and dv = et dt

4.∫

x2 sin 2x dx : Choose u = x2 and v = sin 2x dx

5.

∫ln(x) dx : Choose u = ln x and v = dx

Examples using strategy

1.

∫x sin x dx : Choose u = x and dv = sin x dx

2.

∫xex dx : Choose u = x and dv = ex dx

3.

∫t2et dt : Choose u = t2 and dv = et dt

4.∫

x2 sin 2x dx : Choose u = x2 and v = sin 2x dx

5.

∫ln(x) dx : Choose u = ln x and v = dx

Examples using strategy

1.

∫x sin x dx : Choose u = x and dv = sin x dx

2.

∫xex dx : Choose u = x and dv = ex dx

3.

∫t2et dt : Choose u = t2 and dv = et dt

4.∫

x2 sin 2x dx : Choose u = x2 and v = sin 2x dx

5.

∫ln(x) dx : Choose u = ln x and v = dx

Examples using strategy

1.

∫x sin x dx : Choose u = x and dv = sin x dx

2.

∫xex dx : Choose u = x and dv = ex dx

3.

∫t2et dt : Choose u = t2 and dv = et dt

4.∫

x2 sin 2x dx : Choose u = x2 and v = sin 2x dx

5.

∫ln(x) dx : Choose u = ln x and v = dx