Chapter 6 The Definite Integral. § 6.1 Antidifferentiation.

Chapter 6

The Definite Integral

§ 6.1

Antidifferentiation

Antidifferentiation

Definition Example

Antidifferentiation: The process of determining f (x) given f ΄(x)

If , then xxf 2

.2xxf

Finding Antiderivatives

EXAMPLEEXAMPLE

Find all antiderivatives of the given function.

89xxf

Theorems of Antidifferentiation

The Indefinite Integral

Rules of Integration


EXAMPLEEXAMPLE

Determine the following.

dx

xxx

3

12 2


EXAMPLEEXAMPLE

Find the function f (x) for which and f (1) = 3. xxxf 2

Antiderivatives in Application

EXAMPLEEXAMPLE

A rock is dropped from the top of a 400-foot cliff. Its velocity at time t seconds is v(t) = -32t feet per second.

(a) Find s(t), the height of the rock above the ground at time t.(b) How long will the rock take to reach the ground?

(c) What will be its velocity when it hits the ground?

§ 6.2

Areas and Riemann Sums

Area Under a Graph

Definition Example

Area Under the Graph of f (x) from a to b: An example of this is shown to the right

Area Under a Graph

In this section we will learn to estimate the area under the graph of f (x) from x = a to x = b by dividing up the interval into partitions (or subintervals),

each one having width where n = the number of partitions that

will be constructed. In the example below, n = 4.n

abx

A Riemann Sum is the sum of the areas of the rectangles generated above.

Riemann Sums to Approximate Areas

EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Use a Riemann sum to approximate the area under the graph f (x) on the given interval using midpoints of the subintervals

4,22;2 nxxxf

The partition of -2 ≤ x ≤ 2 with n = 4 is shown below. The length of each subinterval is

.1

4

22

x

-2 2x1 x2 x3 x4

x

x


Observe the first midpoint is units from the left endpoint, and the midpoints themselves are units apart. The first midpoint is x1 = -2 + = -2 + .5 = -1.5. Subsequent midpoints are found by successively adding

CONTINUECONTINUEDD

x.1x

2/x2/x

midpoints: -1.5, -0.5, 0.5, 1.5

The corresponding estimate for the area under the graph of f (x) is

xfxfxfxf 5.15.05.05.1

xffff 5.15.05.05.1

5125.225.025.025.2

So, we estimate the area to be 5 (square units).

Approximating Area With Midpoints of Intervals

CONTINUECONTINUEDD

-0.5

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

-3 -2 -1 0 1 2 3


EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Use a Riemann sum to approximate the area under the graph f (x) on the given interval using left endpoints of the subintervals

5,31;3 nxxxf

The partition of 1 ≤ x ≤ 3 with n = 5 is shown below. The length of each subinterval is

.4.05

13

x

3

x

x1 x2 x3 x4 x5

1 1.4 1.8 2.2 2.6


The corresponding Riemann sum is

CONTINUECONTINUEDD

xfxfxfxfxf 6.22.28.14.11

xfffff 6.22.28.14.11

12.154.06.22.28.14.11 33333

So, we estimate the area to be 15.12 (square units).

Approximating Area Using Left Endpoints

CONTINUECONTINUEDD

0

5

10

15

20

25

30

1 1.4 1.8 2.2 2.6 3 3.4

§ 6.3

Definite Integrals and the Fundamental Theorem


Δx = (b – a)/n, x1, x2, …., xn are selected points from a partition [a, b].

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

0 0.2 0.4 0.6 0.8 1 1.2

Calculating Definite Integrals

EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Calculate the following integral.

1

05.0 dxx

The figure shows the graph of the function f (x) = x + 0.5. Since f (x) is nonnegative for 0 ≤ x ≤ 1, the definite integral of f (x) equals the area of the shaded region in the figure below.

10.5

1


The region consists of a rectangle and a triangle. By geometry,

5.05.01heightwidthrectangle of area

CONTINUECONTINUEDD

5.0112

1heightwidth

2

1 triangleof area

Thus the area under the graph is 0.5 + 0.5 = 1, and hence

.15.01

0 dxx


EXAMPLEEXAMPLE

Calculate the following integral.

1

1xdx

The Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus

EXAMPLEEXAMPLE

Use the Fundamental Theorem of Calculus to calculate the following integral.

1

0

5.031 13 dxex x

Use TI 83 to compute the definite integral: 1) put f(x) into y1 and graph.2) 2nd trace 73) Enter lower limit and upper limit at the prompts.

Area Under a Curve as an Antiderivative

§ 6.4

Areas in the xy-Plane

Properties of Definite Integrals

Area Between Two Curves

Finding the Area Between Two Curves

EXAMPLEEXAMPLE

Find the area of the region between y = x2 – 3x and the x-axis (y = 0) from x = 0 to x = 4.

Finding the Area Between Two Curves

EXAMPLEEXAMPLE

Write down a definite integral or sum of definite integrals that gives the area of the shaded portion of the figure.

§ 6.5

Applications of the Definite Integral

Average Value of a Function Over an Interval


EXAMPLEEXAMPLE

Determine the average value of f (x) = 1 – x over the interval -1 ≤ x ≤ 1.


EXAMPLEEXAMPLE

(Average Temperature) During a certain 12-hour period the temperature at time

t (measured in hours from the start of the period) was degrees. What was the average temperature during that period?

2

3

1447 tt

Consumers’ Surplus

Consumers’ Surplus

EXAMPLEEXAMPLE

Find the consumers’ surplus for the following demand curve at the given sales level x.

20;10

3 xx

p

Chapter 6 The Definite Integral. § 6.1 Antidifferentiation.

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Transcript of Chapter 6 The Definite Integral. § 6.1 Antidifferentiation.