Math Boot Camp

Math Boot CampLecture 8: Calculus: Integration

Matthew Morse

8/31/2018

Math Boot Camp

7: The Integral

7.1: The Definite Integral as a Limit of Sums

7.1: Integration

There are two fundamental questions which calculus addresses.The derivative allows us to answer questions about rates of change.

The integral allows us to answer questions about the area ofregions defined by functions. In addition to the basic application ofcomputations of area, this has applications in fields includingprobability, where probabilities are represented by area andcomputed using integrals. Because the theory of statistics is builton probability, an understanding of integration is necessary tounderstand statistics.

Math Boot Camp

7: The Integral

7.1: The Definite Integral as a Limit of Sums

7.1: Integration

Geometry tells us how to compute the area of simple shapes, likerectangles, triangles, and circles. Integration tells us how tocompute the area of shapes defined by functions.

As a first example, consider the region bounded by the vertical linesx = 2, x = 4, the horizontal x-axis, and the function f (x) = x2.

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7: The Integral

7.1: The Definite Integral as a Limit of Sums

7.1: Region Defined by f (x) = x2

0 1 2 3 4 5

4

8

12

16

20

Rx

x2

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7: The Integral

7.1: The Definite Integral as a Limit of Sums

7.1: Rectangular Approximations of Area

As a first step, we can approximate the area of the region R byusing a rectangle. Three of the borders of the region are straightlines, so we can use them as three sides of the rectangle. Onepossible choice for the fourth side is to draw a horizontal line at anappropriate height.

Then the area of the rectangle is the width of the rectangle, whichis 4− 2 = 2, multiplied by the height of the line.

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7: The Integral

7.1: The Definite Integral as a Limit of Sums

7.1: Two Possible Rectangles

0 1 2 3 4 5

4

8

12

16

20

Rx

x2

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7: The Integral

7.1: The Definite Integral as a Limit of Sums

7.1: Subdividing the Width

The difficulty in using one rectangle to approximate the area of theregion is that the width is large, and so there are many possiblechoices for the value of f (x) to use as the height.

If we subdivide the width of the region into many small pieces,then f (x) varies less over each piece, and the area estimate will bebetter.

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7: The Integral

7.1: The Definite Integral as a Limit of Sums

7.1: Using Many Rectangles

0 1 2 3 4 5

4

8

12

16

20

Rx

x2

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7: The Integral

7.1: The Definite Integral as a Limit of Sums

7.1: Riemann Sums

Suppose we subdivide the area into n rectangles, and the width ofeach rectangle is ∆x (“delta x”). For the ith rectangle, we choosea value xi in that subinterval, and take f (xi ) to be the height ofthe rectangle.The area of one rectangle is f (xi ) ∆x , and the summed area of allof the rectangles is

n∑i=1

f (xi ) ∆x

This is known as a Riemann sum.

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7: The Integral

7.1: The Definite Integral as a Limit of Sums

7.1: Riemann Sums and Definite Integrals

As the number of rectangles, n, increases, and so the rectanglewidth, ∆x , decreases, the Riemann sum becomes a better andbetter approximation of the true area of the region. We canexpress the exact area as a limit.

Area = lim∆x→0

n∑i=1

f (xi ) ∆x

We call this limit a definite integral, and write this integral as∫ b

af (x) dx

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7: The Integral

7.1: The Definite Integral as a Limit of Sums

7.1: Integral Notation

A definite integral is written∫ b

af (x) dx

We say, “the integral from a to b of f of x , dx”. This symbolmeans the area between two vertical lines at x = a and x = b, thehorizontal x-axis, and the function f (x).

a and b are called the bounds of integration. Ordinarily, a is thesmaller of the two numbers. f (x) is called the integrand. For theproblem we have been considering, the left bound a = 2, the rightbound b = 4, and the integrand is the function x2, so we can writethe integral ∫ 4

2x2 dx

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7: The Integral

7.2: Indefinite Integrals and the Fundamental Theorem of Calculus

7.2.1: Antiderivatives

Differentiation operates on a function and produces anotherfunction. (The new function tells you the rate of change of theoriginal function.)

We have discussed inverses in various contexts this week. Thebasic principle of an inverse is that it “undoes” the original action.You might ask if there is an inverse operation to differentiation.

In other words, we know that the derivative of f (x) is f ′(x). If westart with a function f (x), can we identify the function F (x) whichwe can differentiate to get f (x)? If so, we will call F (x) theantiderivative of f (x).

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7: The Integral

7.2: Indefinite Integrals and the Fundamental Theorem of Calculus

7.2.1: Antiderivatives

Differentiation is the act of computing a derivative.Antidifferentiation is the act of computing an antiderivative.

Antidifferentiation and differentiation are inverse operations. Thismeans that if we start with a function, find its antiderivative, andthen compute the derivative of the new function, we get back theoriginal function.

We also want to do this in the other direction. If we start with afunction, compute its derivative, and then compute the newfunction’s antiderivative, we want to get back the original function.

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7: The Integral

7.2: Indefinite Integrals and the Fundamental Theorem of Calculus

7.2.1: Antiderivatives Example

Let f (x) = 2x . What is the antiderivative of f (x)? That is, whichfunction has a derivative equal to 2x?

From our discussion of derivatives, we know that the derivative ofx2 is 2x , so F (x) = x2 is an antiderivative of f (x) = 2x .

In the other direction, the derivative of f (x) = 2x is f ′(x) = 2.What is the antiderivative of f ′(x)?

An antiderivative of f ′(x) = 2 is f (x) = 2x .

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7: The Integral

7.2: Indefinite Integrals and the Fundamental Theorem of Calculus

7.2.1: The General Antiderivative

Let f (x) = 2x + 3. What is the derivative of f (x)? f ′(x) = 2.

What is the antiderivative of f ′(x) = 2?

Well, it should be f (x) = 2x + 3, but on the last slide we said itwas f (x) = 2x . What’s going on?

Recall that the derivative of a constant is 0. This means that if weadd any constant to an antiderivative, it is still an antiderivative.Both 2x + 3 and 2x are antiderivatives of 2.

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7: The Integral

7.2: Indefinite Integrals and the Fundamental Theorem of Calculus

7.2.1: The General Antiderivative

All antiderivatives of f (x) = 2 have the form F (x) = 2x + C ,where C is any constant. We call C the constant of integration.

This is always true. If a function has an antiderivative, we canalways add any constant C to the antiderivative. When we speakof the antiderivative of a function, we must always write in + C .

Example

The antiderivative of f (x) = ex is F (x) = ex + C

For any choice of the constant C , the derivative of F (x) is equal toex .

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7: The Integral

7.2: Indefinite Integrals and the Fundamental Theorem of Calculus

7.2.1: Antiderivatives Examples

Some more examples:1) f (x) = x2 + 100F (x) = 1

3x3 + 100x + C

2) f (x) = −3x−4 + ex

F (x) = x−3 + ex + C

3) f (x) = 1x

F (x) = ln |x |+ C

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7: The Integral

7.2: Indefinite Integrals and the Fundamental Theorem of Calculus

7.2.1: Indefinite Integrals

You may ask why we call C the constant of integration. This isbecause another term for antiderivative is indefinite integral. Thenotation we use is ∫

f (x)dx = F (x)

and we say “the indefinite integral of f of x equals F of x .” Thisstatement has exactly the same meaning as “the antiderivative off of x equals F of x .”

Example ∫2x dx = x2 + C

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7: The Integral

7.2: Indefinite Integrals and the Fundamental Theorem of Calculus

7.2.1: Indefinite Integrals Examples

Rewriting the previous examples with the new notation, we have

1) f (x) = x2 + 100∫f (x) dx =

1

3x3 + 100x + C .

2) f (x) = −3x−4 + ex∫f (x) dx = x−3 + ex + C .

3) f (x) =1

x∫f (x) dx = ln|x |+ C

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7: The Integral

7.2: Indefinite Integrals and the Fundamental Theorem of Calculus

7.2.2: The Fundamental Theorem of Calculus

Why do we use nearly the same notation and name for the definiteintegral and the antiderivative?

The Fundamental Theorem of Calculus

Let f (x) be a function and F (x) be the antiderivative of f (x).Then ∫ b

af (x) dx = F (b)− F (a)

Alternate Notation: ∫ b

af (x) dx = F (x)|ba

Read this as “evaluate F of x at b, and then subtract the value ofF of x at a.”

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7: The Integral

7.2: Indefinite Integrals and the Fundamental Theorem of Calculus

7.2.2: The Fundamental Theorem of Calculus

Intuition:

f (x) is the derivative of F (x). f (x) tells us the rate of change ofF (x) at each point.

F (b)− F (a) is the total change in F (x) between a and b.

A definite integral is a sum. We are summing up the instantaneouschange in F that occurs at each moment as x moves from a to b.

The sum of the instantaneous changes is equal to the total change.

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7: The Integral

7.2: Indefinite Integrals and the Fundamental Theorem of Calculus

7.2.2: Fundamental Theorem: Example

Compute

∫ 4

2x2 dx

The bounds of integration are a = 2 and b = 4. For f (x) = x2,

one antiderivative is F (x) =1

3x3.

By the Fundamental Theorem of Calculus,∫ 4

2x2 dx = F (b)− F (a) =

1

3(4)3 − 1

3(2)3 =

56

3

(What happened to the constant of integration C? Because we aretaking the difference of the antiderivatives, C will always cancel.So we just set it to 0.)

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7: The Integral

7.3: Computing Integrals

7.3: Properties of Integrals

The Fundamental Theorem states∫ b

af (x) dx = F (b)− F (a)

Changing the labels,∫ a

bf (x) dx = F (a)− F (b)

Therefore, ∫ b

af (x) dx = −

∫ a

bf (x) dx

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7: The Integral

7.3: Computing Integrals

7.3: Properties of Intergrals

Similarly, ∫ a

af (x) dx = 0

and ∫ b

af (x) dx =

∫ c

af (x) dx +

∫ b

cf (x) dx

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7: The Integral

7.3: Computing Integrals

7.3.1: Powers

Since integration is a process that undoes differentiation, there aresimilar rules for integration that mirror or “undo” the rules ofdifferentiation. These are written in terms of indefinite integrals.

For example, the power rule for integrals is given by:∫xndx =

xn+1

n + 1+ C

where n 6= −1.

(Proof:d

dx

(xn+1

n + 1

)=

(n + 1)xn+1−1

n + 1= xn)

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7: The Integral

7.3: Computing Integrals

7.3.1: Powers

As stated in a previous example,∫1

xdx =

∫x−1 dx = ln|x |+ C

Therefore we can find the integral of any power.

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7: The Integral

7.3: Computing Integrals

7.3.1: Powers Examples

∫x5 dx =

x6

6+ C

∫1

x3dx =

∫x−3 dx =

x−2

−2+ C = − 1

2x2+ C

∫ √x dx =

∫x1/2 dx =

x3/2

3/2+ C

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7: The Integral

7.3: Computing Integrals

7.3.2: Exponentials

∫ex dx = ex + C

Slightly more complicated,∫ax dx =

ax

ln a+ C

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7: The Integral

7.3: Computing Integrals

7.3.3: Logarithms

∫ln(x) dx = x ln(x)− x + C

Check:

d

dx(x ln x − x) = 1 · ln(x) + x · 1

x− 1 = ln(x) + 1− 1 = ln(x)

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7: The Integral

7.3: Computing Integrals

7.3.5: The Integral Is a Linear Operator

Let f (x) and g(x) be integrable functions and let a and b beconstants. Then∫

(af (x) + bg(x)) dx = a

∫f (x) dx + b

∫g(x) dx

This rule is the counterpart to the summation rule and themultiply by constant rule for derivatives. The derivative is also alinear operator.

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7: The Integral

7.3: Computing Integrals

7.3.5: Linear Rule Examples

1)

∫(60x − 6x2) dx = 60

x2

2− 6

x3

3+ C = 30x2 − 2x3 + C

2)

∫3 ln(x) dx = 3(x ln(x)− x) + C

3)

∫ √t − 1

tdt =

∫(t−1/2 − 1

t)dt =

t1/2

1/2− ln|t|+ C

= 2t1/2 − ln|t|+ C

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7: The Integral

7.3: Computing Integrals

7.3.6: Integration by Substitution

There are two rules for differentiation which we do not yet havecounterparts for. The first rule for differentiation is the chain rule.The equivalent statement for integration is integration bysubstitution. We have been stating these rules as indefiniteintegrals. Integration by Substitution also applies for indefiniteintegrals, but we state it in the form for definite integrals.∫ b

af (g(x))g ′(x) dx =

∫ g(b)

g(a)f (u) du

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7: The Integral

7.3: Computing Integrals

7.3.6: Integration by Substitution

The starting idea is that in the chain rule, we identify an “innerfunction” and replace that function with a new variable, u. (Wedecompose the function into g(u) and f (x), where f (x) = u.)

We want to do the same thing here, but we want to substitute forthe variable of integration at the same time as we replace the innerfunction.

This technique is useful when we have a complicated integrandwhich we cannot directly integrate, but we can identify a potentialinner function to replace.

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7: The Integral

7.3: Computing Integrals

7.3.6: Substitution Example

Evaluate: ∫ 1

02(2x + 1)3 dx

We could find this integral by first expanding the cubic term, andthen distributing the integral over each polynomial termindividually.

However, we recognize 2x + 1 as an inner function, so we can tryintegration by substitution.

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7: The Integral

7.3: Computing Integrals

7.3.6: Substitution Example

With the integrand 2(2x + 1)3, we set u = g(x) = 2x + 1.

The substitution formula is∫ b

af (g(x))g ′(x) dx =

∫ g(b)

g(a)f (u) du

so in order for this to work, we must also identify the functionsf (x) and the derivative of g(x).

g ′(x) = 2. Fortunately the integrand has a coefficient of 2, so allthat remains is to identify f (u). If we set f (u) = u3, then

2(2x + 1)3 = f (g(x))g ′(x)

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7: The Integral

7.3: Computing Integrals

7.3.6: Substitution Example

We have identified the functions f (u) and g(x) to give the left sidethe correct form, so we can make the substitution. If we werefinding an indefinite integral, we would be done. We can write∫

2(2x + 1)3 dx =

∫f (g(x))g ′(x) dx =

∫f (u) du

=

∫u3 du =

u4

4+ C =

(2x + 1)4

4+ C

But when we are computing a definite integral, there is one morestep.

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7: The Integral

7.3: Computing Integrals

7.3.6: Substitution Example

When we change the variables in the integral, we can change thebounds of integration in the same way. Recall that u = g(x) and aand b are values that x takes. So when we replace x with u, wecan replace a and b with g(a) and g(b) at the same time.

In this case, a = 0, b = 1, and g(x) = 2x + 1, so g(0) = 1 andg(1) = 3. So our solution becomes∫ 1

02(2x + 1)3 dx =

∫ g(b)

g(a)f (u) du =

u4

4

∣∣∣∣31

=81

4− 1

4= 20

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7: The Integral

7.3: Computing Integrals

7.3.7: Integration by Parts

Integration by substitution is the counterpart to the chain rule forderivatives.

The counterpart to the product rule for differentiation is calledintegration by parts. The rule is∫

f (x)g ′(x) dx = f (x)g(x)−∫

f ′(x)g(x) dx

The use is similar to integration by substitution, but morecomplicated. You must be able to break down the integrand intothe product of two functions, f (x), and g ′(x) where you know theintegral of g ′(x).

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7: The Integral

7.3: Computing Integrals

7.3.7: Integration by Parts

For example, consider the integral∫xexdx .

We can split this integral up using integration by parts by assigningf (x) = x and g ′(x) = ex . When we do so, note that f ′(x) = 1 andg(x) = ex .

Thus, our integral can be simplified using integration by parts, asfollows:∫

xex dx =

∫f (x)g ′(x) dx = f (x)g(x)−

∫f ′(x)g(x) dx

= xex −∫

ex dx = xex − ex + C

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7: The Integral

7.3: Computing Integrals

7.4: Rules of Integration

Fundamental Thm∫ ba f (x) dx = F (b)− F (a)

Linear rule∫

(af (x) + bg(x)) dx = a∫f (x) dx + b

∫g(x) dx

Power rule 1∫xn dx = xn+1

n+1 + C (n 6= −1)

Power rule 2∫x−1 dx = ln|x |+ C

Exponential 1∫ex dx = ex + C

Exponential 2∫ax dx = ax

ln a + C

Logarithm∫

ln(x) dx = x ln(x)− x + C

Substitution∫ ba f (g(x)g ′(x) dx =

∫ g(b)g(a) f (u) du

Parts∫f (x)g ′(x) dx = f (x)g(x)−

∫f ′(x)g(x) dx