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6.1 The Indefinite IntegralAn Antiderivative of a function f is a function F such that F = f.

X4 is an antiderivative of 4x3

X4 + 45 is also an antiderivative of 4x3

Indefinite integral of a function : set of all antiderivatives of that function

Because the derivative of x4 is 4x3

And the derivative of x4 + 45 is 4x3

f (x)dx : “the indefinite integral of f (x) with respect to x”

the set of all antiderivatives of f.

f is called the integrand, and x is called the variable of integration.

4x3dx = x4 + C Every possible antiderivative of 4x3 has the form x4 + C.

Check that

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Rules for Indefinite IntegralsPower Rule for the Indefinite Integral

Rules for Exponential Functions

Rules for Sums/Differences --- Constant Multiples --- Absolute Value

Examples

(if n ≠ –1)

[f(x) g(x)]dx = f(x)dx g(x)dx

k f(x) dx = k f(x)dx

|x| dx = x |x| + C

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Application: Finding Cost from Marginal Cost

The marginal cost to produce baseball caps at a production level of x caps is 4 – 0.001x dollars per cap, and the cost of producing 100 caps is $500. Find the cost function.

Recall : Marginal cost function is the derivative of the cost function

So, C (x) = 4 – 0.001x Must find C(x).

C(x) = (4 – 0.001x)dx

C(100) = 500

C(100) = 4(100) – 0.0005(100)2 + K

500 = 395 + K

K = 105.

C(x) = 4x – 0.0005x2 + 105

Now we must find K

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Motion in a Straight Line

Position, Velocity, and Acceleration: Integral Form

Example

Velocity of a particle moving in a straight line is given by v(t) = 4t + 1

=> position after t seconds is s(t) = v(t)dt = (4t + 1)dt = 2t2 + t + C.

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Another Example

The velocity of a particle moving along in a straight line is given by

v(t) = 4t + 1 m/s. The particle is at position s = 2 meters at time t = 1. Find an expression for s in terms of t.

s(t) = v(t)dt = (4t + 1)dt S(1) = 2

S(t) = 2t2 + t + C 2 = 2(1)2 + 1 + C

C = -1

Thus, s(t) = 2t2 + t – 1 metersFor a freely falling body experiencing no air resistance and zero initial velocity, find an expression for the velocity v in terms of t. [Note: On Earth, a freely falling body experiencing no air resistance accelerates downward at approximately 9.8 m/s2 (or 32 ft/s2).]

a(t) = –9.8 m/s2

= (–9.8)dt

= –9.8t + C.

v(t) = a(t)dt

V(0) = 0

0 = –9.8(0) + C

C = 0

v(t) = –9.8t m/s

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Motion in a Straight Line

Vertical Motion Under Gravity: Velocity and Position

If one ignores air resistance, the vertical velocity and position of an object moving under gravity are given by

British Units Metric Units

Velocity: v(t) = –32t + v0 ft/s v(t) = –9.8t + v0 m/s

Position: s(t) = –16t 2 + v0t + s0 ft s(t) = –4.9t 2 + v0t + s0 m

v0 = initial velocity = velocity at time 0

s0 = initial position = position at time 0Example

If a ball is thrown down at 2 ft/s from a height of 200 ft, then its velocity and position after t seconds are

v(t) = –32t – 2 ft/s and s(t) = –16t 2 – 2t + 200 ft.

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6.2 Substitution Technique for Integration

The chain rule for derivatives gives us the antiderivative technique.

Recall : to differentiate a function like (x2 + 1)6, we first think of the function

as g(u) where u = x2 + 1 and g(u) = u6. We then compute the derivative,

using the chain rule:

Thus, d/dx (x2 + 1)6 = 6(x2 + 1)5 (2x) = 12x (x2 + 1)5

Suppose we want to evaluate an integral such as: ∫ 4x(x2 + 1)6 dx.

Substitution Method1. Write u as a function of x.

2. Take the derivative du/dx and solve for the quantity dx in terms of du.

3. Use the expression you obtain in step 2 to substitute for dx in the given integral and substitute u for its defining expression.

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Example: – Substitution

Find ∫ 4x(x2 + 1)6 dx.

Step 1: Choose an expression for “u”. Often an expression raised to a power is the best choice. Write u as a function of x.

Step 2: Take the derivative of u with respect to x

Step 3: Solve for dx

Step 4: Substitute u for the defining expression and dx in the original integral:

= ∫ 2u6 du

Step 5: Solve the integral andSubstitute back in for u for the final answer

∫ 4x(x2 + 1)6 dx = ∫ 4xu6 du

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Shortcuts for (ax + b) expressions Rule Quick Example

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When to Use Substitution & What to use for “u”

See FAQs on P. 476 in your Text for an Excellent Resource

In General: Pretend you want to take the derivative instead of the integral.If taking the derivative would require chain rule then integratingMay require substitution.

There are no rules for choosing “u” Common choices for “u” are:

•A linear expression raised to a power: Set u = the linear expression(3x -2)-3 => Let u = 3x – 2

•A constant raised to a linear expression: Set u = the linear expression32x+1 => Let u = 2x + 1

There are several more on page 476. Check them out!

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6.3 The Definite integral

The Indefinite Integral, has many solutions with infinitely many “Constants” of Integration and the integral is not defined for any particular interval.

The Definite Integral IS defined for a particular interval [a, b] and has only one solution - a number.

Graphically: The definite integral gives the area under a curve from [a, b] We form rectangles under the curve, dividing the area into a number of subdivisions, then add the rectangle areas together to estimate the total area under the curve and thus, the definite integral.

Numerically: We use a formula to calculate the sum of the subdivision areas We call this a Riemann Sum. (Note: There are “left” and “right” riemann sums)

Check out this website for some clear pictures showing the difference:http://www2.seminolestate.edu/lvosbury/CalculusI_Folder/RiemannSumDemo.htm

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Example1: Total Cost

A cell phone company’s pricing scheme marginal cost of a phone call is

dollars per hour. Estimate the total cost of

a 2 hour phone call. (Left Riemann Sum Examples)

Note: x = (b – a)/n

Minute-by-Minute (120 subdivisions)Cost ≈ $1.56

Second-by-Second (7,200 subdivisions)Cost ≈ $1.52

5 x 2 = $10 5(1) + .45(1) = $5.45

5(.5) + .83(.5) +.45(.5) + .31(.5)= $3.30

5(.25) + 1.43(.25) +.83(.25) + .59(.25) +… = $2.31

= f (x0)x + f (x1)x +· · ·+ f (xn – 1)x. = (xk)x

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The Definite Integral: Numerical and Graphical Approaches

Interpretation of the Riemann SumIf f is the rate of change of a quantity F (that is, f = F ), then the Riemann sum of f approximates the total change of F from x = a to x = b. (The approximation improves as the number of subdivisions increases toward infinity.)

Interpretation ExamplesIf f (t) is the rate of change in the number of bats in a belfry and [a, b] = [2, 3], then the Riemann sum approximates the total change in the number of bats in the belfry from time t = 2 to time t = 3.

If c(x) is the marginal cost of producing the xth item and [a, b] = [10, 20], then the Riemann sum approximates the cost of producing items 11 through 20.

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The Definite Integral: Numerical and Graphical Approaches

The Definite Integral

If f is a continuous function, the definite integral of f from

a to b is defined to be the limit of the Riemann sums as

the number of subdivisions approaches infinity:

If f is the rate of change of a quantity F (that is, f = F), thenis the (exact) total change of F from x = a to x = b.

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6.4 The Definite Integral: Algebraic Approach

The Fundamental Theorem of Calculus (FTC)

Let f be a continuous function defined on the interval [a, b] and let F be any antiderivative of f defined on [a, b]. Then

Moreover, an antiderivative of f is guaranteed to exist.

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Application: Total Cost

Your cell phone company offers you an innovative pricing scheme. When you make a call, the marginal cost is

dollars per hour.

Compute the total cost of a 2-hour phone call.