Warm-Up

4-1: Antiderivatives & Indefinite Integrals ©2002 Roy L. Gover ( Objectives: Define the antiderivative (indefinite integral) Learn basic antidifferentiation rules Solve simple differential equations

Definition If F(x) is an antiderivative of f(x) on an interval I, then F ’ (x)=f(x) + c where c is a constant.

Example Let be an antiderivative of f(x) then

The Problem... Find the function given its derivative. The function is the antiderivative or indefinite integral.

Example the function whose derivative is. Confirm your answer. If,find f(x),

Try This the function whose derivative is. Confirm your answer. If,find f(x),

Try This the function whose derivative is. Confirm your answer. If,find f(x),

Try This the function whose derivative is. Confirm your answer. If,find f(x),

Try This the function whose derivative is. If,find f(x), aka the power rule for integration.

Example Find the antiderivative of 2 x 2.

Example Find the antiderivative of cos x.

Try This Find the antiderivative of sin x. -cos x+c

Example Find the general solution of the differential equation:

The general solution to: is: y= 4 x+c because:

Multiple solutions exist because: Important Idea

4 Solutions to : y=4x+c

Definition Integrand Variable of Integration Constant of Integration Integral Sign

Example Find the antiderivative:

Power Rule for Integration Special Case : Definition

Sum& Difference Rule for Integration The integral of the sum is the sum of the integrals… Definition The integral of the difference is the difference of the integrals

Example Evaluate the indefinite integral:

Try This Find the antiderivative : Hint: re-write and use the power rule.

Assignment page all,15-27 odd

Assignment page all,15-27 odd

Warm-Up Find the antiderivative : Hint: Re-write as 2 fractions.

Solution

Example Evaluate the indefinite integral:

Lesson Close See page 250 of your text for basic integration rules. Since integration is the inverse of differentiation, you already know most of the rules. Memorize

Intro to Differential Equations. We need to do some work on Differential equations…

Example Sometimes we want a particular antiderivative... Find the equation for y given that y ’=2 x +3 and y passes through the point (2,1)

Example Find the equation for y given that y ’=2 x +3 and y passes through the point (2,1)

You try Find the general solution of F'(x) = x 3 + 2x and the particular solution passing through point (2, 6) Note: (2, 6) means F(2) = 6

Particular Solutions Find the general solution of F'(x) = x 3 + 2x and the particular solution passing through point (2, 6) F(x) = ∫x 3 + 2x = x 4 + x 2 + C 4 F(2) = 6 = C 4 general solution F(x) = x 4 + x particular solution -2 = c

Solving a Problem A ball is thrown upward with an initial velocity of 64 feet per second from an initial height of 80 feet. a) Find the position function giving the height s as a function of time t. b) When does the ball hit the ground?

Assignment page 255 #35-41odd, odd, 63, 69.

QR code activity 1) If you do not have a QR code scanner, partner with someone who does. 2) Choose a random QR Code to scan, and solve the given problem. 3) In the appropriate blank, record your answer and justification on the back of your worksheet. 4) Use that answer to find the next problem. (e.g. The answer to #7 should be at the bottom of the QR code for #8)

Solving a Problem A ball is thrown upward with an initial velocity of 64 feet per second from an initial height of 80 feet. a) Find the position function giving the height s as a function of time t. b) When does the ball hit the ground? s' = ∫s'' = ∫-32 feet/sec position = s velocity = s' acceleration = s'' s' = -32t + c How can we find c?

Solving a Problem A ball is thrown upward with an initial velocity of 64 feet per second from an initial height of 80 feet. a) Find the position function giving the height s as a function of time t. b) When does the ball hit the ground? position = s velocity = s' acceleration = s'' s' = -32t + c initial velocity is 64 when time is zero 64 = -32(0) + c 64 = c s' = -32t + 64

Solving a Problem A ball is thrown upward with an initial velocity of 64 feet per second from an initial height of 80 feet. a) Find the position function giving the height s as a function of time t. b) When does the ball hit the ground? position = s velocity = s' acceleration = s'' s = ∫s' = ∫-32t + 64 s = -32t t + c 2 How do we find c this time?

Solving a Problem A ball is thrown upward with an initial velocity of 64 feet per second from an initial height of 80 feet. a) Find the position function giving the height s as a function of time t. b) When does the ball hit the ground? position = s velocity = s' acceleration = s'' s = -32t t + c 2 initial height is 80 (time is 0) 80 = -32(0) (0) + c 2 80 = c

Solving a Problem A ball is thrown upward with an initial velocity of 64 feet per second from an initial height of 80 feet. a) Find the position function giving the height s as a function of time t. b) When does the ball hit the ground? position = s velocity = s' acceleration = s'' 80 = -32(0) (0) + c 2 80 = c s = -16t 2 +64t + 80 So when does the ball hit the ground?

Solving a Problem A ball is thrown upward with an initial velocity of 64 feet per second from an initial height of 80 feet. a) Find the position function giving the height s as a function of time t. b) When does the ball hit the ground? s = -16t 2 +64t + 80 The position will be zero. 0 = -16t 2 +64t = -16(t - 5)(t + 1) t = 5 or -1