Antiderivatives and Initial Value Problems

Transcription

1 Antierivatives an Initial Value Problems Warm up If f (x) =2x, whatisf (x)? Can you think of another function that f (x) coul be? If f (x) =3x 2 +, what is f (x)? Can you think of another function that f (x) coul be?

2 Definition An antierivative of a function f on an interval I is another function F such that F 0 (x) =f (x) forallx 2 I. Examples:. An antierivative of f (x) =2x is F (x) =x Another antierivative of f (x) =2x is F (x) =x There are lots of antierivatives of f (x) = 2x which look like F (x) =x 2 + C. Suppose that h is i erentiable in an interval I, an h 0 (x) =0forallx in I. Then h is a constant function! i.e. h(x) =C for all x 2 I,whereC is a constant. So, if F (x) is one antierivative of f (x), then any other antierivative must be of the form F (x)+c. Example: All of the antierivatives of f (x) = 2x look like for some constant C. F (x) =x 2 + C

3 Every function f that has at least one antierivative F has infinitely many antierivatives We refer to F (x)+c as the F (x)+c. general antierivative or the inefinite integral an enote it by F (x)+c = f (x). Example: 2x = x 2 + C. Examples x 2 = 3 x 3 + C, because ( 3 x 3 + C) = 3 3x 2 = x 2 x 3 = 4 x 4 + C, because ( 4 x 4 + C) = 4 4x 3 = x 3 x 5 = x 3 = x k =

4 Some important basic integrals x k = k + x k+ + C if k 6= x = sin(x) = cos(x) = e x = sec 2 (x) = p x 2 = Theorem (Opposite of sum an constant rules) Suppose the functions f an g both have antierivatives on the interval I. Then for any constants a an b, the function af + bg has an antierivative on I an a f (x)+b g(x) = a f (x) + b g(x)

5 Di erential equations A i erential equation is an equation involving erivatives. The goal is usually to solve for y. Just like you coul use algebra to solve y 2 + x 2 = for y, you can use calculus (an algebra) to solve things like y 5y =0 fory. A solution to a i erential equation is a function you can plug in that satisfies the equation. For example, y = e 5x is a solution to the i erential equation above since e5x =5e 5x, so y 5y =(5e 5x ) 5(e 5x )=0 X. Simplest i erential equations: antierivatives Fining an antierivative can also be thought of as solving a i erential equation: Solve the i erential equation y = x 2. Answer: y = x 2 = 3 x 3 + C. Check: 3 x 3 + C. = 3 3 x 2 +0=x 2 X

6 Examples () Solve the i erential equation y 0 =2x +sin(x). (2) Check that cos(x)+sin(x) is a solution to 2 y 2 + y =0.

7 Initial value problems Fin a solution to the i erential equation also satisfies y (2) = 8/3. general solution: y = x 2 + which y = 3 x 3 + x + C Each color correspons to a choice of C. Re cuve is the particular solution. Initial value problems Fin a solution to the i erential equation also satisfies y (2) = 8/3. general solution: particular solution: y = x 2 + which y = 3 x 3 + x + C y = 3 x 3 + x 2 Each color correspons to a choice of C. Re cuve is the particular solution.

8 Definition An initial-value problem is a i erential equation together with enough aitional conitions to specify the constants of integration that appear in the general solution. The particular solution of the problem is then a function that satisfies both the i erential equation an also the aitional conitions. Solve the initial value problem subject to y(0) = 0. y =2x +sin(x) general solution: y = x 2 cos(x)+c Algebraically: get a particular solution by solving 0 = y(0) = (0) 2 cos(0) + C = +C (for C) C =, so y = x 2 cos(x)+.

9 Solve the initial-value problem y 00 = cos x, y 0 ( 2 )=2, y( 2 )=3. Step : Calculate the antierivative of cos(x) to fin the general solution for y 0. Step 2: Plug in the values y 0 ( 2 )=2tocalculateC. Step 3: Write own the particular solution for y 0. Step 4: Calculate the antierivative of your particular solution in Step 3 to fin the general solution for y. Step 5: Plug in the values y( 2 )=3 to solve for the new constant. Step 6: Write own the particular solution for y.

10 Wor problem: An object roppe from a cli has acceleration a = 9.8 m/sec 2 uner the influence of gravity. What is the function s(t) that moels its height at time t? Initial value problem: Solve 2 s t 2 = 9.8, s(0) = s 0, s 0 (0) = 0.

11 Wor problem: Suppose that a baseball is thrown upwar from the roof of a 00 meter high builing. It hits the street below eight secons later. What was the initial velocity of the baseball, an how high i it rise above the street before beginning its escent? Initial value problem: Solve 2 s = t2 9.8, s(0) = 00, s(8) = 0. Use your solution to () calculate s 0 (0), an (2) solve s 0 (t )=0fort an calculate s(t ).

12 Extra practice: Basic antierivatives Problem A. Inefinite integrals.. x 7 0. (2 5x)(3 + 2x)( x) 2. x 7. px(ax 2 + bx + c) 3. x 2. (x 2 /x 2 ) 3 4. x 5/3 3. ( p x / p x) 5. x 5/4 4. px + px p x2 4p x ( + 2x) 3 x 4 ( + x) 3 p x x 2 (8 x +2x 3 6/x 3 +2x 5 +5x ) 7. 2x 2 + x 2 x 2 8. If f = x /x2 an f() = /2 finf(x). Problem B. Inefinite integrals with trigonometric functions sinx cot x sin x 7 cos x tan 2 x sec x(sec x + tan x) csc x(csc x cot x) 6 cos 2 x + 2 sin 2 x + cot2 x tan x cos x + 2 cos 2 x sinx cos 2 x 3 cos x +4 sin 2 x cos x [hint: mult an ivie by ( + cos(x)) an rewrite using csc(x), etc.] 5. (tan x + cot x) cos x

13 0. tan x sec x + tan x 5. p cos 2x csc x csc x cot x cos x + cos x sin x sin x cos 2x cos 2x p+sin2x 4. p + cos 2x 9. sin 3 x + cos 3 x sin 2 x cos 2 x Problem C. Integrals with exponential functions an inverse functions.. 2 x x x 2 [hint: a an subtract in the numerator] 2. (6x 5 2x 4 7x +3/x 5+4e x +7 x ) x 2 7. x (x/a + a/x + x a + a x + ax) x 6 8. px 3 4. p x 4 + 3p 7 x 2 + 6e x + x x 2 2 p x x p x 2 x 4 +x ax 0. cos (sin x)