Block 5 Stochastic & Dynamic Systems Lesson 14 - Integral ...

Block 5 Stochastic & Dynamic Systems Lesson 14 - Integral ...

Block 5 Stochastic & Dynamic Systems Lesson 14 Integral Calculus The World is now a nonlinear, dynamic, and uncertain place. Block 5 - The End of the Road Lesson 14 - Integral Calculus Lesson 15 Stochastic (Probability) Models Lesson 16 Differential Equations Lesson 17 Dynamic Models f ( x)dx Integral Calculus on a silver platter Integration If F(x) is a function whose derivative F(x) = f(x), then F(x) is called the integral of f(x) For example, F(x) = x3 is an integral of f(x) = 3x2 Note also that G(x) = x3 + 5 and H(x) = x3 6 are also integrals of f(x) I like to call F(x) the antiderivative. Indefinite Integral

The indefinite integral of f(x), denoted by f ( x)dx F ( x) C where C is an arbitrary constant is the most general integral of f(x) The indefinite integral of f(x) = 3x2 is 2 3 3x dx x C A Strategy Gosh, it seems so simple. f ( x)dx F ( x) C the integrand First I guess at a function whose derivative is f(x) and then I add a constant of integration.

or use a table of integrals The top five n 1 x 1. x n dx C ; n 1 n 1 1 2. dx ln x C x 3. a dx ax C 4. e ax ax e dx C a n a bx 5. a bx dx

n 1 b n C Basic Rules of Integration 1. c f ( x)dx c f ( x)dx 2. f ( x) g ( x) dx f ( x) dx g ( x) dx 3. af ( x) bg ( x) dx a f ( x) dx b g ( x) dx The top four and the basic rules in action 10 2 .2 x 1.2 7 x 5e x 2 x 25 dx 1 2 .2 x 7 x dx 5e dx 10 dx 2x 1.2 dx 25dx x 3 .2 x .2 7 x 5e 2x

10 ln x 25 x C 3 .2 .2 Initial Conditions The rate at which annual income (y) changes with respect to years of education (x) is given by dy 100 x3/ 2 ; 4 x 16 dx where y = 28,720 when x = 9. Find y. y 100 x 3/ 2 dx 100 28, 720 40 9 5/ 2 x 3/ 2 1 3/ 2 1

C 40 x 5 / 2 C C 9720 C C 19, 000 y 40 x5/ 2 19, 000 Integrating au , a > 0 u a du ? let a eln a ln a u a du e u du e au e using e du C a au ln a u

ln a u e du C ln a Tricks of the Trade Its Magic Use some algebra 2 4 3 2 2 y y 2 y 2

3 y y dy y 3 3 dy 4 9 C 2 x 1 x 3 dx 1 3 2 x 5 x x 2 2 x 5 x 3 dx C 6 9 12 2 6

x 1 dx x x x 3 2 3 2 2 1 x 1 2 2 dx x x dx C x 2 x Adjusting for du method of substitution 1/ 2 1 2 x x 5 dx 2 2 x x 5 dx 2

1/ 2 1 1 x 5 2 x 5 2 xdx 2 2 3/ 2 2 u u du n 1 n 3/ 2 1 2 x 5 C 3 du u x 5 and 2 x dx 2 n 1 3/ 2

More dus 3 x2 3xe dx 2 e 2 xdx 2 du where u x ; 2 x dx 3 u 3 u e du e 2 2 3 x2 e C 2 x2 2x 1 x 2 5 dx x 2 5 2 xdx du 2 where u x 5; 2 x dx 1 2

du ln u ln x 5 C u Integration by Parts derived from the product rule for derivatives Slick Harry is trying to sell an ENM student a simple integration formula. u dv uv v du x xe dx ? x x

let u x; du dx and dv e dx; v e dx e x x x x x x xe dx xe e dx xe e e x 1 C x Another one?

u dv uv v du ln y dy ? dy let u ln y ; du and v y; dv dy y dy ln y dy ln y y y y y ln y y C y ln y 1 C Integration by Tables A favorite integration formula of engineering students is: dx x 1 px ln a be a be px a pa dx x 1 .1 x find :

ln 5 2 e C .1 x 5 2e 5 .1 5 There is no shame in using a Table of Integrals. Check it out! .1 x 2 e .1 d x 1

1 .1 x ln 5 2e C .1 x dx 5 .1 5 5 .5 5 2 e .5 5 2e .1x 5 .2e .1x .5 5 5 2e .1x 2.5 1 .1 x .1 x 5

2 e 2.5 5 2 e dx x 1 .1 x find : ln 5 2 e C .1 x 5 2e

5 .1 5 Another Table Problem find 7 x 2 ln 4 x dx n 1 n 1 u ln u u Table : u n ln u du C 2 n 1 n 1 set n 2, u 4 x, du 4dx 7 2 7 x ln 4 x dx 43 4 x ln 4 x 4dx 3 3 7 4 x ln 4 x 4 x 1 3 ln 4 x

C C 7 x 64 3 9 9 3 2 An Engineers Favorite Table ax e ax 1. xe dx 2 ax 1 a dx ex 2. x ln 1 e 1 ex 3. ln x dx x ln x x 2 2 x

x 4. x ln x dx ln x 2 4 no, this one The motivated student may wish to verify the above by showing that the derivatives of the right hand functions results in the corresponding integrands. The Definite Integral b f ( x)dx F (b) F (a) where F '( x) f ( x) a Areas under the curve Definite Integral Given a function f(x) that is continuous on the interval [a,b] we divide the interval into n subintervals of equal width, x, and from each interval choose a point, xi*. Then the definite integral of f(x) from a to b is Area under the curve f(x)

x x The Fundamental Theorem of Calculus Let f be a continuous real-valued function defined on a closed interval [a, b]. Let F be a function such that for all x in [a, b] then . Several engineering management students informally meeting after class to discuss the implications of the fundamental theorem Fundamental Theorem Evaluating a definite integral 2 2 4 x dx F (2) F (0) 0 4 23 3

32 0 3 3 4 x where F ( x) 4 x 2 dx 3 2 x 1 2 x 1 dx 0 33 1 26 3 3 3 3

3 2 0 2 1 3 3 0 1 3 3 The Area under a curve The area under the curve of a probability density function over its entire domain is always equal to one. Verify that the following function is a probability density function: 2 3t f (t ) 9 , 0 t 1000 10

1000 2 3 3t t dt 9 9 10 10 0 1000 0 3 3 10 10 9

1 Area between Curves Find the area bounded by y = 4 4x2 and y = x2 - 1 y1 = 4 4x2 y1 - y2 = 5 5x2 20 (5 5 x ) dx 1 3 1 (-1, 0) (1, 0) y 2 = x2 - 1 Improper Integrals r

f ( x) dx lim f ( x) dx a r a b b f ( x) dx lim f ( x) dx r 0 r f ( x) dx f ( x) dx f ( x) dx

0 and if the limit exists, the integral is said to be convergent. Otherwise it is divergent, right? Example an Improper Integral r 2 r 1 1 x dx lim 3 dx lim 3 r x

r x 2 1 1 1 1 1 1 lim 2 0 r 2r 2 2 2 1 Lets do another one r ke 0 x x

dx lim ke dx lim ke r 0 r lim ke k k r An engineering professor caught doing some improper integration. r x r 0 The Engineers Little Table of Improper Definite Integrals 1. e ax 0

1 dx a 2. xe ax 0 1 dx 2 a n ax 3. x e 0 n! dx n 1 a

4. x e dx 4 0 2 x2 A little table for engineers Some Applications Taking it to the limit The Crime Rate The total number of crimes is increasing at the rate of 8t + 10 where t = months from the start of the year. How many crimes will be committed during the last 6 months of the year? given : d crime dt 8t 10 12 8t 10 dt 4t

6 2 10t 12 6 4 144 120 4 36 60 492 Learning Curves Cumulative Cost Y (i ) ai b x hours to produce ith unit x T ( x) Y (i ) ai i 1 T ( x) V ( x) x

i 1 b cumulative direct labor hrs to produce x units average unit hours to produce x units Learning Curves Approximate Cumulative Cost x x b 1 ax T ( x) Y (i ) di a i di b 1 0 0 b b 1 b

ax ax V ( x) (b 1) x b 1 Learning Curves - example Production of the first 10 F-222s, the Air Forces new steam driven fighter, resulted in a 71 percent learning curve in dollar cost where the first aircraft cost $18 million. What will be cost of the second lot of 10 aircraft? (sim-lc 18e6 20 71) 20 .5059 20 T ( x) 18 i 10 .4941 18 x di .5059 10

18 .5059 .5059 20 10 $47.903 million .5059 ln(71/100) 47.903 b V ( x) $4.790 million ln 2 10 .4941 The Average of a Function The average or mean value of a function y = f(x) over the interval [a,b] is given by: b 1 y f ( x)dx

b a a Find the average of the function y = x2 over the interval [1,3]: 3 3 3 1 x 27 1 26 1 2 y x dx 4 3 1 1 6 6 3 3 2 1 Average profit An oil companys profit in dollars for the qth million gallons sold is given by P = P(q) = 369q 2.1q2 400 If the company sells 100 million gallons this year,

what is the average profit per gallon sold? 100 369q 2.1q 0 100 2 400 dq 2 3 100 1 369q 2.1q 400q 100 2 3 0

1 369(104 ) 2.1(106 ) 2 400(10 ) 100 2 3 104 369 210 0 4 3 100 2 $11,050 100 184.5 74 11, 050 or $.01105 6 10 An Inventory Problem

Demand for an item is constant over time at the rate of 720 per year. Whenever the on-hand inventory reaches zero, a shipment of 60 units is received. The inventory holding cost is based upon the average on-hand inventory. Let y = 60 720t be the on-hand inventory as a function of time where t is in years. It takes 60/720 = 1/12 yr to go from an inventory of 60 to 0. 1 y 1/12 0 1/12 1/12 60 720t 2 60 720t dt 12 60t 2 0 0 2

60 60 1 12 6 720 60 30 2 12 12 t Annuities Let A = the present value of a continuous annuity at an annual rate r (compounded continuously) for T years if a payment at time t is at the rate of f(t) per year. Then T rt A f (t )e dt 0 A is the present value of a continuous income stream Annuity Example Determine the present value of a continuous annuity at an annual rate of 8% for 10 years if the payment at

time t is at the rate of 1000t dollars per year. 10 A (1000t )e 10 .08 t 0 dt 1000 te .08 t dt 1000 0 e .08(10) 1 1000 .08(10) 1 2 2 .08 .08 1000 126.37377 156.25 $29,876 ax

e ax xe dx a 2 ax 1 e 10 .08t .08 2 .08t 1 0 Annuities Let S = the accumulated amount of a continuous annuity at an annual rate r (compounded continuously) for T years if a payment at time t is at the rate of f(t) per year. Then T S f (t )e 0

r (T t ) dt Back to the example 10 10 .08(10 t ) S 1000t e 0 .8 dt 1000e te .08t dt 0 10

e .08t 2225.541 .08t 1 2 .08 0 e .8 1 2225.541 .8 1 2 2 .08 .08 2225.541 126.3737 156.25 $66, 491 More of that darn example Recall continuous compounding S Pe rt .08 10 S $29,876e

$66, 490 Iterated Integrals Evaluate 1 1 0 0 2 3 3 y x dxdy 1 x2 0 0 ( x 2 xy y 2 )dydx

Double Integral Evaluate the integral over R where R is the triangle formed by y = x, y = 0, x = 1. 2 1 x 0 0 2 2 1 1 2 2 ( x y ) dA ( x y

) dydx ( x y )dxdy R 2 0 y

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