Introduction To Integrals

A. Rectangular Approximation Methdods (RAM)

Describe the drip rate Describe the amount spent per month

How long does the tank drip?

What is the area under the curve? What is the area under the curve?

How much liquid drips out? What is the total amount spent?

The ________________ is equal to the summation of a rate over an interval of the dependant

variable !

If we cannot easily find the area under the curve we can use a Rectangular Approximation Method:

- divide region in intervals of length Δx

- consider rectangles of length Δx and height f(x)

- you must decide where in the interval Δx to chose the height

LRAM - left side of interval

RRAM - right side of interval

MRAM - middle of interval

LRAM RRAM

MRAM

Trapezoidal Approximation 1) works just like rectangle approximation (often more efficient) 2) f(x) at right and f(x) at left ends of interval are the bases of the trapezoid, and Δx is the height 3) area = 1/2 (b1 + b2) h

1. LRAM – 4 intervals

2 .RRAM – 4 intervals

3. MRAM – 4 intervals

4. Trapezoidal Method – 4 intervals

5. Area under the x-axis counts as negative LRAM – 7 intervals: 0-1, 1-2, 2-3, 3-4, 4-4.5, 4.5-5, 5-6

6. Area under the x-axis counts as negative RRAM – 7 intervals: 0-1, 1-2, 2-3, 3-4, 4-4.5, 4.5-5, 5-6

7. MRAM and Trapezoidal – 5 intervals

8. 1997 AB Exam Question #9 A bug begins to crawl up a vertical wire at time t = 0. The velocity v of the bug at time t,

0 8x , is given by the function whose graph is shown above. What is the total distance the bug traveled from t = 0 to t = 8?

(A) 14 (B) 13 (C) 11 (D) 8 (E) 6

9. Given a a semicircle centered at the origin with a radius of 2,

a) Estimate the area of the semicircle using MRAM and 4 intervals b) Repeat using the trapezoidal method and 4 intervals. c) Compare your answer to the actual area

x

y

10. Estimate the area under the sine curve on the interval [0, π] using RRAM and 4 intervals MRAM and 4 intervals LRAM and 4 intervals Trapezoidal method and 4 intervals How would your answer’s differ if the interval were [0, 2π]?

Approximatation Methods Questions

1. Accuracy of RAM increases as the number of intervals __________________ 2. ________ overestimates an increasing function 3. ________ overestimates a decreasing function 4. ________ underestimates an increasing function 5. ________ underestimates an increasing function 6. When calculating the net area under a curve, area below the x-axis counts as ______________ 7. ___________ is often the most accurate method 8. True or False: the intervals must always be equally spaced when using RAM ______ 9. The formula for the area of a trapezoid is ______________________ 10 . When would you want to have many closely spaced intervals, and when would it make more sense to have fewer widely spaced intervals

Presenting the Integral

1. Any of the Reimann Sum methods of approximating the area under a curve* will approach the

exact area if we let the size of the biggest interval approach zero.

The integral of f(x) from a to b is defined to be the area under the curve:

f (x)dx

a

b

ò = limn®¥

f (ck )Dxkk=1

n

å

- a and b are the lower and upper bounds

- f(x) is the integran, x is the variable of integration

- the function is divided into n subintervals

- the width of the kth subinterval is

Dxk and ck is a

value of the function anywhere on that subinterval

2. If f(x) is continuous then it is integrable. If f(x) is not continuous, then it may or may

not be integrable

Our definition of integrable here means that we can in theory find the area under the

curve. In practice, many functions can only have their integral approximated by

numerical means .

2. Net Area under curve:

If the curve lies both above and below the axis, the integral gives:

Area above – Area below the x-axis

Sketch the sine curve on [0,

2p]. Can you guess what

sin(x)dx

0

2p

ò equals?

* Area under the curve means “net” area between the curve and the x-axis. Area above the x-axis is positive, area below the x-axis is negative.

Graph the function and evaluate the integral:

Y = 3 on [0,4]

3dx

0

4

ò y = x on [0,4]

x dx

0

4

ò

y = ½ x – 1 on [-2,6]

(1

2x - 3) dx

-2

6

ò

y = 25 - x2 [-5,5]

25 - x2 dx

-5

5

ò

The function f is continuous on the closed interval 2, 8 and has values that are given in the table

above. Using the subintervals 2, 5 , 5, 7 , 7, 8 , what is the trapezoidal approximation of

8

2f x dx

(A) 110 (B) 130 (C) 160 (D) 190 (E) 210

x 2 5 10 14

f x 12 28 34 30

The function f is continuous on the closed interval 2,14 and has values as shown in the table

above. Using the subintervals 2,5 , 5,10 ,and 10,14 ,what is the approximation of 14

2f x dx

found by using a right Riemann sum? (A) 296 (B) 312 (C) 343 (D) 374 (E) 390

x 2 5 7 8 f(x) 10 30 40 20

We can consider the area under the curve peicewise.This lets us integrate peicewise continuous

functions.

Given f(x) shown below, Given f(x) shown below,

evaluate

f (x) dx

-4

3

ò and

f (x) dx

-3

-4

ò

Given: 1

02)( dxxf

2

13)( dxxf

1

01)( dxxg

2

04)( dxxg

1. 2

0)( dxxf

2. 2

0)(2)( dxxgxf

3. 2

1)( dxxg 4.

0

2)( dxxg

5. 1

2)(3 dxxf

6. 1

1)( dxxg

7. 2

0)(3)(2 dxxgxf

8. 1

04)(3)(2 dxxgxf

9. 0

2

2

1)()( dxxfdxxf