Lesson 4-10b

Anti-Differentiation

Quiz

Estimate the area under the graph of f(x) = x² + 1 from x = -1 to x = 2 …. Improve your estimate by using six right endpoint rectangles.

Objectives

• Understand the concept of an antiderivative

• Understand the geometry of the antiderivative and that of slope fields

• Work rectilinear motion problems with antiderivatives

Vocabulary

• Antiderivative – the opposite of the derivative, if f(x) = F’(x) then F(x) is the antiderivative of f(x)

• Integrand – what is being taken the integral of [F’(x)]

• Variable of integration – what variable we are taking the integral with respect to

• Constant of integration – a constant (derivative of which would be zero) that represents the family of functions that could have the same derivative

Two other Anti-derivative Forms

1. Form:

2. Form:

F(x) = ex dx = ex + C ∫

1F(x) = ----- dx = ln |x| + C x∫

Remember derivative of ex is just ex

Remember derivative of ln x is (1/x)

Practice Problems

a) (2ex + 1) dx∫

b) (1 – x-1) dx∫

-5c) ----- dx x∫

d) (ex + x² - 1) dx∫

2ex + x + C

ex + ⅓x3 - x + Cx – ln |x| + C

-5 ln|x| + C

How to Find C

• In order to find the specific value of the constant of integration, we need to have an initial condition to evaluate the function at (to solve for C)!

• Example: find such that F(1) = 4.F(x) = (2x + 3) dx∫

F(x) = x² + 3x + C

F(1) = 4 = (1)² + 3(1) + C = 4 + C 0 = C (boring answer!)

Acceleration, Velocity, Position

• Remember the following equation from Physics:

s(t) = s0 + v0t – ½ at²

where s0 is the initial offset distance (when t=0) v0 is the initial velocity (when t=0) and a is the acceleration constant (due to gravity)

• We can solve problems given either s(t) or a(t) (and some initial conditions) – basically solving the problem from either direction!

Motion Problems

• Find the velocity function v(t) and position function s(t) corresponding to the acceleration function a(t) = 4t + 4 given v(0) = 8 and s(0) = 12.

v(t) = (4t + 4) dt = 2t² + 4t + v0∫v(0) = 8 = 2(0)² + 4(0) + v0 8 = v0

= 2t² + 4t + 8

s(t) = (2t² + 4t + 8) dt = ⅔t³ + 2t² + 8t + s0∫s(0) = 12 = ⅔(0)³ + 2(0)² + 8(0) + s0

12 = s0

s(t) = ⅔t³ + 2t² + 8t + 12

Motion Problems

• A ball is dropped from a window hits the ground in 5 seconds. How high is the window (in feet)?

v(t) = a(t) dt = -32t + v0∫v(0) = 0 = -32(0) + v0 (ball was dropped) 0 = v0

s(t) = v(t) dt = -16t² + s0∫s(5) = 0 = -16(5)² + s0

s0 = 400 feet window was 400 feet up

a(t) = - 32 ft/s²

Slope Fields

• A slope field is the slope of the tangent to F(x) (f(x) in an anti-differentiation problem) plotted at each value of x and y in field.

• Since the constant of integration is unknown, we get a family of curves.

• An initial condition allows us to plot the function F(x) based on the slope field.

y

x

f(0) = -2

Slope Field Example

Summary & Homework

• Summary:– Anti-differentiation is the reverse of the derivative– It introduces the integral– One of its main applications is area under the

curve

• Homework: – pg 358-360: Day 1: 1-3, 12, 13, 16

Day 2: 25, 26, 53, 61, 74